1132 lines
29 KiB
Markdown
1132 lines
29 KiB
Markdown
<!-- GFM-TOC -->
|
||
* [递归](#递归)
|
||
* [1. 树的高度](#1-树的高度)
|
||
* [2. 平衡树](#2-平衡树)
|
||
* [3. 两节点的最长路径](#3-两节点的最长路径)
|
||
* [4. 翻转树](#4-翻转树)
|
||
* [5. 归并两棵树](#5-归并两棵树)
|
||
* [6. 判断路径和是否等于一个数](#6-判断路径和是否等于一个数)
|
||
* [7. 统计路径和等于一个数的路径数量](#7-统计路径和等于一个数的路径数量)
|
||
* [8. 子树](#8-子树)
|
||
* [9. 树的对称](#9-树的对称)
|
||
* [10. 最小路径](#10-最小路径)
|
||
* [11. 统计左叶子节点的和](#11-统计左叶子节点的和)
|
||
* [12. 相同节点值的最大路径长度](#12-相同节点值的最大路径长度)
|
||
* [13. 间隔遍历](#13-间隔遍历)
|
||
* [14. 找出二叉树中第二小的节点](#14-找出二叉树中第二小的节点)
|
||
* [层次遍历](#层次遍历)
|
||
* [1. 一棵树每层节点的平均数](#1-一棵树每层节点的平均数)
|
||
* [2. 得到左下角的节点](#2-得到左下角的节点)
|
||
* [前中后序遍历](#前中后序遍历)
|
||
* [1. 非递归实现二叉树的前序遍历](#1-非递归实现二叉树的前序遍历)
|
||
* [2. 非递归实现二叉树的后序遍历](#2-非递归实现二叉树的后序遍历)
|
||
* [3. 非递归实现二叉树的中序遍历](#3-非递归实现二叉树的中序遍历)
|
||
* [BST](#bst)
|
||
* [1. 修剪二叉查找树](#1-修剪二叉查找树)
|
||
* [2. 寻找二叉查找树的第 k 个元素](#2-寻找二叉查找树的第-k-个元素)
|
||
* [3. 把二叉查找树每个节点的值都加上比它大的节点的值](#3-把二叉查找树每个节点的值都加上比它大的节点的值)
|
||
* [4. 二叉查找树的最近公共祖先](#4-二叉查找树的最近公共祖先)
|
||
* [5. 二叉树的最近公共祖先](#5-二叉树的最近公共祖先)
|
||
* [6. 从有序数组中构造二叉查找树](#6-从有序数组中构造二叉查找树)
|
||
* [7. 根据有序链表构造平衡的二叉查找树](#7-根据有序链表构造平衡的二叉查找树)
|
||
* [8. 在二叉查找树中寻找两个节点，使它们的和为一个给定值](#8-在二叉查找树中寻找两个节点，使它们的和为一个给定值)
|
||
* [9. 在二叉查找树中查找两个节点之差的最小绝对值](#9-在二叉查找树中查找两个节点之差的最小绝对值)
|
||
* [10. 寻找二叉查找树中出现次数最多的值](#10-寻找二叉查找树中出现次数最多的值)
|
||
* [Trie](#trie)
|
||
* [1. 实现一个 Trie](#1-实现一个-trie)
|
||
* [2. 实现一个 Trie，用来求前缀和](#2-实现一个-trie，用来求前缀和)
|
||
<!-- GFM-TOC -->
|
||
|
||
|
||
# 递归
|
||
|
||
一棵树要么是空树，要么有两个指针，每个指针指向一棵树。树是一种递归结构，很多树的问题可以使用递归来处理。
|
||
|
||
## 1. 树的高度
|
||
|
||
[104. Maximum Depth of Binary Tree (Easy)](https://leetcode.com/problems/maximum-depth-of-binary-tree/description/)
|
||
|
||
```java
|
||
public int maxDepth(TreeNode root) {
|
||
if (root == null) return 0;
|
||
return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
|
||
}
|
||
```
|
||
|
||
## 2. 平衡树
|
||
|
||
[110. Balanced Binary Tree (Easy)](https://leetcode.com/problems/balanced-binary-tree/description/)
|
||
|
||
```html
|
||
3
|
||
/ \
|
||
9 20
|
||
/ \
|
||
15 7
|
||
```
|
||
|
||
平衡树左右子树高度差都小于等于 1
|
||
|
||
```java
|
||
private boolean result = true;
|
||
|
||
public boolean isBalanced(TreeNode root) {
|
||
maxDepth(root);
|
||
return result;
|
||
}
|
||
|
||
public int maxDepth(TreeNode root) {
|
||
if (root == null) return 0;
|
||
int l = maxDepth(root.left);
|
||
int r = maxDepth(root.right);
|
||
if (Math.abs(l - r) > 1) result = false;
|
||
return 1 + Math.max(l, r);
|
||
}
|
||
```
|
||
|
||
## 3. 两节点的最长路径
|
||
|
||
[543. Diameter of Binary Tree (Easy)](https://leetcode.com/problems/diameter-of-binary-tree/description/)
|
||
|
||
```html
|
||
Input:
|
||
|
||
1
|
||
/ \
|
||
2 3
|
||
/ \
|
||
4 5
|
||
|
||
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
|
||
```
|
||
|
||
```java
|
||
private int max = 0;
|
||
|
||
public int diameterOfBinaryTree(TreeNode root) {
|
||
depth(root);
|
||
return max;
|
||
}
|
||
|
||
private int depth(TreeNode root) {
|
||
if (root == null) return 0;
|
||
int leftDepth = depth(root.left);
|
||
int rightDepth = depth(root.right);
|
||
max = Math.max(max, leftDepth + rightDepth);
|
||
return Math.max(leftDepth, rightDepth) + 1;
|
||
}
|
||
```
|
||
|
||
## 4. 翻转树
|
||
|
||
[226. Invert Binary Tree (Easy)](https://leetcode.com/problems/invert-binary-tree/description/)
|
||
|
||
```java
|
||
public TreeNode invertTree(TreeNode root) {
|
||
if (root == null) return null;
|
||
TreeNode left = root.left; // 后面的操作会改变 left 指针，因此先保存下来
|
||
root.left = invertTree(root.right);
|
||
root.right = invertTree(left);
|
||
return root;
|
||
}
|
||
```
|
||
|
||
## 5. 归并两棵树
|
||
|
||
[617. Merge Two Binary Trees (Easy)](https://leetcode.com/problems/merge-two-binary-trees/description/)
|
||
|
||
```html
|
||
Input:
|
||
Tree 1 Tree 2
|
||
1 2
|
||
/ \ / \
|
||
3 2 1 3
|
||
/ \ \
|
||
5 4 7
|
||
|
||
Output:
|
||
3
|
||
/ \
|
||
4 5
|
||
/ \ \
|
||
5 4 7
|
||
```
|
||
|
||
```java
|
||
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
|
||
if (t1 == null && t2 == null) return null;
|
||
if (t1 == null) return t2;
|
||
if (t2 == null) return t1;
|
||
TreeNode root = new TreeNode(t1.val + t2.val);
|
||
root.left = mergeTrees(t1.left, t2.left);
|
||
root.right = mergeTrees(t1.right, t2.right);
|
||
return root;
|
||
}
|
||
```
|
||
|
||
## 6. 判断路径和是否等于一个数
|
||
|
||
[Leetcdoe : 112. Path Sum (Easy)](https://leetcode.com/problems/path-sum/description/)
|
||
|
||
```html
|
||
Given the below binary tree and sum = 22,
|
||
|
||
5
|
||
/ \
|
||
4 8
|
||
/ / \
|
||
11 13 4
|
||
/ \ \
|
||
7 2 1
|
||
|
||
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
|
||
```
|
||
|
||
路径和定义为从 root 到 leaf 的所有节点的和。
|
||
|
||
```java
|
||
public boolean hasPathSum(TreeNode root, int sum) {
|
||
if (root == null) return false;
|
||
if (root.left == null && root.right == null && root.val == sum) return true;
|
||
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
|
||
}
|
||
```
|
||
|
||
## 7. 统计路径和等于一个数的路径数量
|
||
|
||
[437. Path Sum III (Easy)](https://leetcode.com/problems/path-sum-iii/description/)
|
||
|
||
```html
|
||
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
|
||
|
||
10
|
||
/ \
|
||
5 -3
|
||
/ \ \
|
||
3 2 11
|
||
/ \ \
|
||
3 -2 1
|
||
|
||
Return 3. The paths that sum to 8 are:
|
||
|
||
1. 5 -> 3
|
||
2. 5 -> 2 -> 1
|
||
3. -3 -> 11
|
||
```
|
||
|
||
路径不一定以 root 开头，也不一定以 leaf 结尾，但是必须连续。
|
||
|
||
```java
|
||
public int pathSum(TreeNode root, int sum) {
|
||
if (root == null) return 0;
|
||
int ret = pathSumStartWithRoot(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
|
||
return ret;
|
||
}
|
||
|
||
private int pathSumStartWithRoot(TreeNode root, int sum) {
|
||
if (root == null) return 0;
|
||
int ret = 0;
|
||
if (root.val == sum) ret++;
|
||
ret += pathSumStartWithRoot(root.left, sum - root.val) + pathSumStartWithRoot(root.right, sum - root.val);
|
||
return ret;
|
||
}
|
||
```
|
||
|
||
## 8. 子树
|
||
|
||
[572. Subtree of Another Tree (Easy)](https://leetcode.com/problems/subtree-of-another-tree/description/)
|
||
|
||
```html
|
||
Given tree s:
|
||
3
|
||
/ \
|
||
4 5
|
||
/ \
|
||
1 2
|
||
|
||
Given tree t:
|
||
4
|
||
/ \
|
||
1 2
|
||
|
||
Return true, because t has the same structure and node values with a subtree of s.
|
||
|
||
Given tree s:
|
||
|
||
3
|
||
/ \
|
||
4 5
|
||
/ \
|
||
1 2
|
||
/
|
||
0
|
||
|
||
Given tree t:
|
||
4
|
||
/ \
|
||
1 2
|
||
|
||
Return false.
|
||
```
|
||
|
||
```java
|
||
public boolean isSubtree(TreeNode s, TreeNode t) {
|
||
if (s == null) return false;
|
||
return isSubtreeWithRoot(s, t) || isSubtree(s.left, t) || isSubtree(s.right, t);
|
||
}
|
||
|
||
private boolean isSubtreeWithRoot(TreeNode s, TreeNode t) {
|
||
if (t == null && s == null) return true;
|
||
if (t == null || s == null) return false;
|
||
if (t.val != s.val) return false;
|
||
return isSubtreeWithRoot(s.left, t.left) && isSubtreeWithRoot(s.right, t.right);
|
||
}
|
||
```
|
||
|
||
## 9. 树的对称
|
||
|
||
[101. Symmetric Tree (Easy)](https://leetcode.com/problems/symmetric-tree/description/)
|
||
|
||
```html
|
||
1
|
||
/ \
|
||
2 2
|
||
/ \ / \
|
||
3 4 4 3
|
||
```
|
||
|
||
```java
|
||
public boolean isSymmetric(TreeNode root) {
|
||
if (root == null) return true;
|
||
return isSymmetric(root.left, root.right);
|
||
}
|
||
|
||
private boolean isSymmetric(TreeNode t1, TreeNode t2) {
|
||
if (t1 == null && t2 == null) return true;
|
||
if (t1 == null || t2 == null) return false;
|
||
if (t1.val != t2.val) return false;
|
||
return isSymmetric(t1.left, t2.right) && isSymmetric(t1.right, t2.left);
|
||
}
|
||
```
|
||
|
||
## 10. 最小路径
|
||
|
||
[111. Minimum Depth of Binary Tree (Easy)](https://leetcode.com/problems/minimum-depth-of-binary-tree/description/)
|
||
|
||
树的根节点到叶子节点的最小路径长度
|
||
|
||
```java
|
||
public int minDepth(TreeNode root) {
|
||
if (root == null) return 0;
|
||
int left = minDepth(root.left);
|
||
int right = minDepth(root.right);
|
||
if (left == 0 || right == 0) return left + right + 1;
|
||
return Math.min(left, right) + 1;
|
||
}
|
||
```
|
||
|
||
## 11. 统计左叶子节点的和
|
||
|
||
[404. Sum of Left Leaves (Easy)](https://leetcode.com/problems/sum-of-left-leaves/description/)
|
||
|
||
```html
|
||
3
|
||
/ \
|
||
9 20
|
||
/ \
|
||
15 7
|
||
|
||
There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
|
||
```
|
||
|
||
```java
|
||
public int sumOfLeftLeaves(TreeNode root) {
|
||
if (root == null) return 0;
|
||
if (isLeaf(root.left)) return root.left.val + sumOfLeftLeaves(root.right);
|
||
return sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right);
|
||
}
|
||
|
||
private boolean isLeaf(TreeNode node){
|
||
if (node == null) return false;
|
||
return node.left == null && node.right == null;
|
||
}
|
||
```
|
||
|
||
## 12. 相同节点值的最大路径长度
|
||
|
||
[687. Longest Univalue Path (Easy)](https://leetcode.com/problems/longest-univalue-path/)
|
||
|
||
```html
|
||
1
|
||
/ \
|
||
4 5
|
||
/ \ \
|
||
4 4 5
|
||
|
||
Output : 2
|
||
```
|
||
|
||
```java
|
||
private int path = 0;
|
||
|
||
public int longestUnivaluePath(TreeNode root) {
|
||
dfs(root);
|
||
return path;
|
||
}
|
||
|
||
private int dfs(TreeNode root){
|
||
if (root == null) return 0;
|
||
int left = dfs(root.left);
|
||
int right = dfs(root.right);
|
||
int leftPath = root.left != null && root.left.val == root.val ? left + 1 : 0;
|
||
int rightPath = root.right != null && root.right.val == root.val ? right + 1 : 0;
|
||
path = Math.max(path, leftPath + rightPath);
|
||
return Math.max(leftPath, rightPath);
|
||
}
|
||
```
|
||
|
||
## 13. 间隔遍历
|
||
|
||
[337. House Robber III (Medium)](https://leetcode.com/problems/house-robber-iii/description/)
|
||
|
||
```html
|
||
3
|
||
/ \
|
||
2 3
|
||
\ \
|
||
3 1
|
||
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
|
||
```
|
||
|
||
```java
|
||
public int rob(TreeNode root) {
|
||
if (root == null) return 0;
|
||
int val1 = root.val;
|
||
if (root.left != null) val1 += rob(root.left.left) + rob(root.left.right);
|
||
if (root.right != null) val1 += rob(root.right.left) + rob(root.right.right);
|
||
int val2 = rob(root.left) + rob(root.right);
|
||
return Math.max(val1, val2);
|
||
}
|
||
```
|
||
|
||
## 14. 找出二叉树中第二小的节点
|
||
|
||
[671. Second Minimum Node In a Binary Tree (Easy)](https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/description/)
|
||
|
||
```html
|
||
Input:
|
||
2
|
||
/ \
|
||
2 5
|
||
/ \
|
||
5 7
|
||
|
||
Output: 5
|
||
```
|
||
|
||
一个节点要么具有 0 个或 2 个子节点，如果有子节点，那么根节点是最小的节点。
|
||
|
||
```java
|
||
public int findSecondMinimumValue(TreeNode root) {
|
||
if (root == null) return -1;
|
||
if (root.left == null && root.right == null) return -1;
|
||
int leftVal = root.left.val;
|
||
int rightVal = root.right.val;
|
||
if (leftVal == root.val) leftVal = findSecondMinimumValue(root.left);
|
||
if (rightVal == root.val) rightVal = findSecondMinimumValue(root.right);
|
||
if (leftVal != -1 && rightVal != -1) return Math.min(leftVal, rightVal);
|
||
if (leftVal != -1) return leftVal;
|
||
return rightVal;
|
||
}
|
||
```
|
||
|
||
# 层次遍历
|
||
|
||
使用 BFS 进行层次遍历。不需要使用两个队列来分别存储当前层的节点和下一层的节点，因为在开始遍历一层的节点时，当前队列中的节点数就是当前层的节点数，只要控制遍历这么多节点数，就能保证这次遍历的都是当前层的节点。
|
||
|
||
## 1. 一棵树每层节点的平均数
|
||
|
||
[637. Average of Levels in Binary Tree (Easy)](https://leetcode.com/problems/average-of-levels-in-binary-tree/description/)
|
||
|
||
```java
|
||
public List<Double> averageOfLevels(TreeNode root) {
|
||
List<Double> ret = new ArrayList<>();
|
||
if (root == null) return ret;
|
||
Queue<TreeNode> queue = new LinkedList<>();
|
||
queue.add(root);
|
||
while (!queue.isEmpty()) {
|
||
int cnt = queue.size();
|
||
double sum = 0;
|
||
for (int i = 0; i < cnt; i++) {
|
||
TreeNode node = queue.poll();
|
||
sum += node.val;
|
||
if (node.left != null) queue.add(node.left);
|
||
if (node.right != null) queue.add(node.right);
|
||
}
|
||
ret.add(sum / cnt);
|
||
}
|
||
return ret;
|
||
}
|
||
```
|
||
|
||
## 2. 得到左下角的节点
|
||
|
||
[513. Find Bottom Left Tree Value (Easy)](https://leetcode.com/problems/find-bottom-left-tree-value/description/)
|
||
|
||
```html
|
||
Input:
|
||
|
||
1
|
||
/ \
|
||
2 3
|
||
/ / \
|
||
4 5 6
|
||
/
|
||
7
|
||
|
||
Output:
|
||
7
|
||
```
|
||
|
||
```java
|
||
public int findBottomLeftValue(TreeNode root) {
|
||
Queue<TreeNode> queue = new LinkedList<>();
|
||
queue.add(root);
|
||
while (!queue.isEmpty()) {
|
||
root = queue.poll();
|
||
if (root.right != null) queue.add(root.right);
|
||
if (root.left != null) queue.add(root.left);
|
||
}
|
||
return root.val;
|
||
}
|
||
```
|
||
|
||
# 前中后序遍历
|
||
|
||
```html
|
||
1
|
||
/ \
|
||
2 3
|
||
/ \ \
|
||
4 5 6
|
||
```
|
||
|
||
- 层次遍历顺序：[1 2 3 4 5 6]
|
||
- 前序遍历顺序：[1 2 4 5 3 6]
|
||
- 中序遍历顺序：[4 2 5 1 3 6]
|
||
- 后序遍历顺序：[4 5 2 6 3 1]
|
||
|
||
层次遍历使用 BFS 实现，利用的就是 BFS 一层一层遍历的特性；而前序、中序、后序遍历利用了 DFS 实现。
|
||
|
||
前序、中序、后序遍只是在对节点访问的顺序有一点不同，其它都相同。
|
||
|
||
① 前序
|
||
|
||
```java
|
||
void dfs(TreeNode root) {
|
||
visit(root);
|
||
dfs(root.left);
|
||
dfs(root.right);
|
||
}
|
||
```
|
||
|
||
② 中序
|
||
|
||
```java
|
||
void dfs(TreeNode root) {
|
||
dfs(root.left);
|
||
visit(root);
|
||
dfs(root.right);
|
||
}
|
||
```
|
||
|
||
③ 后序
|
||
|
||
```java
|
||
void dfs(TreeNode root) {
|
||
dfs(root.left);
|
||
dfs(root.right);
|
||
visit(root);
|
||
}
|
||
```
|
||
|
||
## 1. 非递归实现二叉树的前序遍历
|
||
|
||
[144. Binary Tree Preorder Traversal (Medium)](https://leetcode.com/problems/binary-tree-preorder-traversal/description/)
|
||
|
||
```java
|
||
public List<Integer> preorderTraversal(TreeNode root) {
|
||
List<Integer> ret = new ArrayList<>();
|
||
Stack<TreeNode> stack = new Stack<>();
|
||
stack.push(root);
|
||
while (!stack.isEmpty()) {
|
||
TreeNode node = stack.pop();
|
||
if (node == null) continue;
|
||
ret.add(node.val);
|
||
stack.push(node.right); // 先右后左，保证左子树先遍历
|
||
stack.push(node.left);
|
||
}
|
||
return ret;
|
||
}
|
||
```
|
||
|
||
## 2. 非递归实现二叉树的后序遍历
|
||
|
||
[145. Binary Tree Postorder Traversal (Medium)](https://leetcode.com/problems/binary-tree-postorder-traversal/description/)
|
||
|
||
前序遍历为 root -> left -> right，后序遍历为 left -> right -> root。可以修改前序遍历成为 root -> right -> left，那么这个顺序就和后序遍历正好相反。
|
||
|
||
```java
|
||
public List<Integer> postorderTraversal(TreeNode root) {
|
||
List<Integer> ret = new ArrayList<>();
|
||
Stack<TreeNode> stack = new Stack<>();
|
||
stack.push(root);
|
||
while (!stack.isEmpty()) {
|
||
TreeNode node = stack.pop();
|
||
if (node == null) continue;
|
||
ret.add(node.val);
|
||
stack.push(node.left);
|
||
stack.push(node.right);
|
||
}
|
||
Collections.reverse(ret);
|
||
return ret;
|
||
}
|
||
```
|
||
|
||
## 3. 非递归实现二叉树的中序遍历
|
||
|
||
[94. Binary Tree Inorder Traversal (Medium)](https://leetcode.com/problems/binary-tree-inorder-traversal/description/)
|
||
|
||
```java
|
||
public List<Integer> inorderTraversal(TreeNode root) {
|
||
List<Integer> ret = new ArrayList<>();
|
||
if (root == null) return ret;
|
||
Stack<TreeNode> stack = new Stack<>();
|
||
TreeNode cur = root;
|
||
while (cur != null || !stack.isEmpty()) {
|
||
while (cur != null) {
|
||
stack.push(cur);
|
||
cur = cur.left;
|
||
}
|
||
TreeNode node = stack.pop();
|
||
ret.add(node.val);
|
||
cur = node.right;
|
||
}
|
||
return ret;
|
||
}
|
||
```
|
||
|
||
# BST
|
||
|
||
二叉查找树（BST）：根节点大于等于左子树所有节点，小于等于右子树所有节点。
|
||
|
||
二叉查找树中序遍历有序。
|
||
|
||
## 1. 修剪二叉查找树
|
||
|
||
[669. Trim a Binary Search Tree (Easy)](https://leetcode.com/problems/trim-a-binary-search-tree/description/)
|
||
|
||
```html
|
||
Input:
|
||
|
||
3
|
||
/ \
|
||
0 4
|
||
\
|
||
2
|
||
/
|
||
1
|
||
|
||
L = 1
|
||
R = 3
|
||
|
||
Output:
|
||
|
||
3
|
||
/
|
||
2
|
||
/
|
||
1
|
||
```
|
||
|
||
题目描述：只保留值在 L \~ R 之间的节点
|
||
|
||
```java
|
||
public TreeNode trimBST(TreeNode root, int L, int R) {
|
||
if (root == null) return null;
|
||
if (root.val > R) return trimBST(root.left, L, R);
|
||
if (root.val < L) return trimBST(root.right, L, R);
|
||
root.left = trimBST(root.left, L, R);
|
||
root.right = trimBST(root.right, L, R);
|
||
return root;
|
||
}
|
||
```
|
||
|
||
## 2. 寻找二叉查找树的第 k 个元素
|
||
|
||
[230. Kth Smallest Element in a BST (Medium)](https://leetcode.com/problems/kth-smallest-element-in-a-bst/description/)
|
||
|
||
|
||
中序遍历解法：
|
||
|
||
```java
|
||
private int cnt = 0;
|
||
private int val;
|
||
|
||
public int kthSmallest(TreeNode root, int k) {
|
||
inOrder(root, k);
|
||
return val;
|
||
}
|
||
|
||
private void inOrder(TreeNode node, int k) {
|
||
if (node == null) return;
|
||
inOrder(node.left, k);
|
||
cnt++;
|
||
if (cnt == k) {
|
||
val = node.val;
|
||
return;
|
||
}
|
||
inOrder(node.right, k);
|
||
}
|
||
```
|
||
|
||
递归解法：
|
||
|
||
```java
|
||
public int kthSmallest(TreeNode root, int k) {
|
||
int leftCnt = count(root.left);
|
||
if (leftCnt == k - 1) return root.val;
|
||
if (leftCnt > k - 1) return kthSmallest(root.left, k);
|
||
return kthSmallest(root.right, k - leftCnt - 1);
|
||
}
|
||
|
||
private int count(TreeNode node) {
|
||
if (node == null) return 0;
|
||
return 1 + count(node.left) + count(node.right);
|
||
}
|
||
```
|
||
|
||
## 3. 把二叉查找树每个节点的值都加上比它大的节点的值
|
||
|
||
[Convert BST to Greater Tree (Easy)](https://leetcode.com/problems/convert-bst-to-greater-tree/description/)
|
||
|
||
```html
|
||
Input: The root of a Binary Search Tree like this:
|
||
|
||
5
|
||
/ \
|
||
2 13
|
||
|
||
Output: The root of a Greater Tree like this:
|
||
|
||
18
|
||
/ \
|
||
20 13
|
||
```
|
||
|
||
先遍历右子树。
|
||
|
||
```java
|
||
private int sum = 0;
|
||
|
||
public TreeNode convertBST(TreeNode root) {
|
||
traver(root);
|
||
return root;
|
||
}
|
||
|
||
private void traver(TreeNode node) {
|
||
if (node == null) return;
|
||
traver(node.right);
|
||
sum += node.val;
|
||
node.val = sum;
|
||
traver(node.left);
|
||
}
|
||
```
|
||
|
||
## 4. 二叉查找树的最近公共祖先
|
||
|
||
[235. Lowest Common Ancestor of a Binary Search Tree (Easy)](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/)
|
||
|
||
```html
|
||
_______6______
|
||
/ \
|
||
___2__ ___8__
|
||
/ \ / \
|
||
0 4 7 9
|
||
/ \
|
||
3 5
|
||
|
||
For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
|
||
```
|
||
|
||
```java
|
||
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
|
||
if (root.val > p.val && root.val > q.val) return lowestCommonAncestor(root.left, p, q);
|
||
if (root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p, q);
|
||
return root;
|
||
}
|
||
```
|
||
|
||
## 5. 二叉树的最近公共祖先
|
||
|
||
[236. Lowest Common Ancestor of a Binary Tree (Medium) ](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/)
|
||
|
||
```html
|
||
_______3______
|
||
/ \
|
||
___5__ ___1__
|
||
/ \ / \
|
||
6 2 0 8
|
||
/ \
|
||
7 4
|
||
|
||
For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
|
||
```
|
||
|
||
```java
|
||
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
|
||
if (root == null || root == p || root == q) return root;
|
||
TreeNode left = lowestCommonAncestor(root.left, p, q);
|
||
TreeNode right = lowestCommonAncestor(root.right, p, q);
|
||
return left == null ? right : right == null ? left : root;
|
||
}
|
||
```
|
||
|
||
## 6. 从有序数组中构造二叉查找树
|
||
|
||
[108. Convert Sorted Array to Binary Search Tree (Easy)](https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/description/)
|
||
|
||
```java
|
||
public TreeNode sortedArrayToBST(int[] nums) {
|
||
return toBST(nums, 0, nums.length - 1);
|
||
}
|
||
|
||
private TreeNode toBST(int[] nums, int sIdx, int eIdx){
|
||
if (sIdx > eIdx) return null;
|
||
int mIdx = (sIdx + eIdx) / 2;
|
||
TreeNode root = new TreeNode(nums[mIdx]);
|
||
root.left = toBST(nums, sIdx, mIdx - 1);
|
||
root.right = toBST(nums, mIdx + 1, eIdx);
|
||
return root;
|
||
}
|
||
```
|
||
|
||
## 7. 根据有序链表构造平衡的二叉查找树
|
||
|
||
[109. Convert Sorted List to Binary Search Tree (Medium)](https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/description/)
|
||
|
||
```html
|
||
Given the sorted linked list: [-10,-3,0,5,9],
|
||
|
||
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
|
||
|
||
0
|
||
/ \
|
||
-3 9
|
||
/ /
|
||
-10 5
|
||
```
|
||
|
||
```java
|
||
public TreeNode sortedListToBST(ListNode head) {
|
||
if (head == null) return null;
|
||
if (head.next == null) return new TreeNode(head.val);
|
||
ListNode preMid = preMid(head);
|
||
ListNode mid = preMid.next;
|
||
preMid.next = null; // 断开链表
|
||
TreeNode t = new TreeNode(mid.val);
|
||
t.left = sortedListToBST(head);
|
||
t.right = sortedListToBST(mid.next);
|
||
return t;
|
||
}
|
||
|
||
private ListNode preMid(ListNode head) {
|
||
ListNode slow = head, fast = head.next;
|
||
ListNode pre = head;
|
||
while (fast != null && fast.next != null) {
|
||
pre = slow;
|
||
slow = slow.next;
|
||
fast = fast.next.next;
|
||
}
|
||
return pre;
|
||
}
|
||
```
|
||
|
||
## 8. 在二叉查找树中寻找两个节点，使它们的和为一个给定值
|
||
|
||
[653. Two Sum IV - Input is a BST (Easy)](https://leetcode.com/problems/two-sum-iv-input-is-a-bst/description/)
|
||
|
||
```html
|
||
Input:
|
||
|
||
5
|
||
/ \
|
||
3 6
|
||
/ \ \
|
||
2 4 7
|
||
|
||
Target = 9
|
||
|
||
Output: True
|
||
```
|
||
|
||
使用中序遍历得到有序数组之后，再利用双指针对数组进行查找。
|
||
|
||
应该注意到，这一题不能用分别在左右子树两部分来处理这种思想，因为两个待求的节点可能分别在左右子树中。
|
||
|
||
```java
|
||
public boolean findTarget(TreeNode root, int k) {
|
||
List<Integer> nums = new ArrayList<>();
|
||
inOrder(root, nums);
|
||
int i = 0, j = nums.size() - 1;
|
||
while (i < j) {
|
||
int sum = nums.get(i) + nums.get(j);
|
||
if (sum == k) return true;
|
||
if (sum < k) i++;
|
||
else j--;
|
||
}
|
||
return false;
|
||
}
|
||
|
||
private void inOrder(TreeNode root, List<Integer> nums) {
|
||
if (root == null) return;
|
||
inOrder(root.left, nums);
|
||
nums.add(root.val);
|
||
inOrder(root.right, nums);
|
||
}
|
||
```
|
||
|
||
## 9. 在二叉查找树中查找两个节点之差的最小绝对值
|
||
|
||
[530. Minimum Absolute Difference in BST (Easy)](https://leetcode.com/problems/minimum-absolute-difference-in-bst/description/)
|
||
|
||
```html
|
||
Input:
|
||
|
||
1
|
||
\
|
||
3
|
||
/
|
||
2
|
||
|
||
Output:
|
||
|
||
1
|
||
```
|
||
|
||
利用二叉查找树的中序遍历为有序的性质，计算中序遍历中临近的两个节点之差的绝对值，取最小值。
|
||
|
||
```java
|
||
private int minDiff = Integer.MAX_VALUE;
|
||
private TreeNode preNode = null;
|
||
|
||
public int getMinimumDifference(TreeNode root) {
|
||
inOrder(root);
|
||
return minDiff;
|
||
}
|
||
|
||
private void inOrder(TreeNode node) {
|
||
if (node == null) return;
|
||
inOrder(node.left);
|
||
if (preNode != null) minDiff = Math.min(minDiff, node.val - preNode.val);
|
||
preNode = node;
|
||
inOrder(node.right);
|
||
}
|
||
```
|
||
|
||
## 10. 寻找二叉查找树中出现次数最多的值
|
||
|
||
[501. Find Mode in Binary Search Tree (Easy)](https://leetcode.com/problems/find-mode-in-binary-search-tree/description/)
|
||
|
||
```html
|
||
1
|
||
\
|
||
2
|
||
/
|
||
2
|
||
|
||
return [2].
|
||
```
|
||
|
||
答案可能不止一个，也就是有多个值出现的次数一样多。
|
||
|
||
```java
|
||
private int curCnt = 1;
|
||
private int maxCnt = 1;
|
||
private TreeNode preNode = null;
|
||
|
||
public int[] findMode(TreeNode root) {
|
||
List<Integer> maxCntNums = new ArrayList<>();
|
||
inOrder(root, maxCntNums);
|
||
int[] ret = new int[maxCntNums.size()];
|
||
int idx = 0;
|
||
for (int num : maxCntNums) {
|
||
ret[idx++] = num;
|
||
}
|
||
return ret;
|
||
}
|
||
|
||
private void inOrder(TreeNode node, List<Integer> nums) {
|
||
if (node == null) return;
|
||
inOrder(node.left, nums);
|
||
if (preNode != null) {
|
||
if (preNode.val == node.val) curCnt++;
|
||
else curCnt = 1;
|
||
}
|
||
if (curCnt > maxCnt) {
|
||
maxCnt = curCnt;
|
||
nums.clear();
|
||
nums.add(node.val);
|
||
} else if (curCnt == maxCnt) {
|
||
nums.add(node.val);
|
||
}
|
||
preNode = node;
|
||
inOrder(node.right, nums);
|
||
}
|
||
```
|
||
|
||
# Trie
|
||
|
||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/5c638d59-d4ae-4ba4-ad44-80bdc30f38dd.jpg"/> </div><br>
|
||
|
||
Trie，又称前缀树或字典树，用于判断字符串是否存在或者是否具有某种字符串前缀。
|
||
|
||
## 1. 实现一个 Trie
|
||
|
||
[208. Implement Trie (Prefix Tree) (Medium)](https://leetcode.com/problems/implement-trie-prefix-tree/description/)
|
||
|
||
```java
|
||
class Trie {
|
||
|
||
private class Node {
|
||
Node[] childs = new Node[26];
|
||
boolean isLeaf;
|
||
}
|
||
|
||
private Node root = new Node();
|
||
|
||
public Trie() {
|
||
}
|
||
|
||
public void insert(String word) {
|
||
insert(word, root);
|
||
}
|
||
|
||
private void insert(String word, Node node) {
|
||
if (node == null) return;
|
||
if (word.length() == 0) {
|
||
node.isLeaf = true;
|
||
return;
|
||
}
|
||
int index = indexForChar(word.charAt(0));
|
||
if (node.childs[index] == null) {
|
||
node.childs[index] = new Node();
|
||
}
|
||
insert(word.substring(1), node.childs[index]);
|
||
}
|
||
|
||
public boolean search(String word) {
|
||
return search(word, root);
|
||
}
|
||
|
||
private boolean search(String word, Node node) {
|
||
if (node == null) return false;
|
||
if (word.length() == 0) return node.isLeaf;
|
||
int index = indexForChar(word.charAt(0));
|
||
return search(word.substring(1), node.childs[index]);
|
||
}
|
||
|
||
public boolean startsWith(String prefix) {
|
||
return startWith(prefix, root);
|
||
}
|
||
|
||
private boolean startWith(String prefix, Node node) {
|
||
if (node == null) return false;
|
||
if (prefix.length() == 0) return true;
|
||
int index = indexForChar(prefix.charAt(0));
|
||
return startWith(prefix.substring(1), node.childs[index]);
|
||
}
|
||
|
||
private int indexForChar(char c) {
|
||
return c - 'a';
|
||
}
|
||
}
|
||
```
|
||
|
||
## 2. 实现一个 Trie，用来求前缀和
|
||
|
||
[677. Map Sum Pairs (Medium)](https://leetcode.com/problems/map-sum-pairs/description/)
|
||
|
||
```html
|
||
Input: insert("apple", 3), Output: Null
|
||
Input: sum("ap"), Output: 3
|
||
Input: insert("app", 2), Output: Null
|
||
Input: sum("ap"), Output: 5
|
||
```
|
||
|
||
```java
|
||
class MapSum {
|
||
|
||
private class Node {
|
||
Node[] child = new Node[26];
|
||
int value;
|
||
}
|
||
|
||
private Node root = new Node();
|
||
|
||
public MapSum() {
|
||
|
||
}
|
||
|
||
public void insert(String key, int val) {
|
||
insert(key, root, val);
|
||
}
|
||
|
||
private void insert(String key, Node node, int val) {
|
||
if (node == null) return;
|
||
if (key.length() == 0) {
|
||
node.value = val;
|
||
return;
|
||
}
|
||
int index = indexForChar(key.charAt(0));
|
||
if (node.child[index] == null) {
|
||
node.child[index] = new Node();
|
||
}
|
||
insert(key.substring(1), node.child[index], val);
|
||
}
|
||
|
||
public int sum(String prefix) {
|
||
return sum(prefix, root);
|
||
}
|
||
|
||
private int sum(String prefix, Node node) {
|
||
if (node == null) return 0;
|
||
if (prefix.length() != 0) {
|
||
int index = indexForChar(prefix.charAt(0));
|
||
return sum(prefix.substring(1), node.child[index]);
|
||
}
|
||
int sum = node.value;
|
||
for (Node child : node.child) {
|
||
sum += sum(prefix, child);
|
||
}
|
||
return sum;
|
||
}
|
||
|
||
private int indexForChar(char c) {
|
||
return c - 'a';
|
||
}
|
||
}
|
||
```
|
||
|
||
|
||
|
||
|
||
|
||
# 微信公众号
|
||
|
||
|
||
更多精彩内容将发布在微信公众号 CyC2018 上，你也可以在公众号后台和我交流学习和求职相关的问题。另外，公众号提供了该项目的 PDF 等离线阅读版本，后台回复 "下载" 即可领取。公众号也提供了一份技术面试复习大纲，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点，后台回复 "大纲" 即可领取。我基本是按照这个大纲来进行复习的，对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。
|
||
|
||
|
||
<div align="center"><img width="580px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/other/公众号海报2.png"></img></div>
|