28 KiB
28 KiB
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。
树的高度
104. Maximum Depth of Binary Tree (Easy)
public int maxDepth(TreeNode root) {
if (root == null) return 0;
return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
}
平衡树
110. Balanced Binary Tree (Easy)
3
/ \
9 20
/ \
15 7
平衡树左右子树高度差都小于等于 1
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);
}
两节点的最长路径
543. Diameter of Binary Tree (Easy)
Input:
1
/ \
2 3
/ \
4 5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
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;
}
翻转树
226. Invert Binary Tree (Easy)
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;
}
归并两棵树
617. Merge Two Binary Trees (Easy)
Input:
Tree 1 Tree 2
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
Output:
3
/ \
4 5
/ \ \
5 4 7
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;
}
判断路径和是否等于一个数
Leetcdoe : 112. Path Sum (Easy)
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。
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);
}
统计路径和等于一个数的路径数量
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。
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;
}
子树
572. Subtree of Another Tree (Easy)
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.
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);
}
树的对称
1
/ \
2 2
/ \ / \
3 4 4 3
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);
}
最小路径
111. Minimum Depth of Binary Tree (Easy)
树的根节点到叶子节点的最小路径长度
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;
}
统计左叶子节点的和
404. Sum of Left Leaves (Easy)
3
/ \
9 20
/ \
15 7
There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
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;
}
相同节点值的最大路径长度
687. Longest Univalue Path (Easy)
1
/ \
4 5
/ \ \
4 4 5
Output : 2
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);
}
间隔遍历
337. House Robber III (Medium)
3
/ \
2 3
\ \
3 1
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
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);
}
找出二叉树中第二小的节点
671. Second Minimum Node In a Binary Tree (Easy)
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。
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。
一棵树每层节点的平均数
637. Average of Levels in Binary Tree (Easy)
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;
}
得到左下角的节点
513. Find Bottom Left Tree Value (Easy)
Input:
1
/ \
2 3
/ / \
4 5 6
/
7
Output:
7
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
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
void dfs(TreeNode root) {
visit(root);
dfs(root.left);
dfs(root.right);
}
② 中序
void dfs(TreeNode root) {
dfs(root.left);
visit(root);
dfs(root.right);
}
③ 后序
void dfs(TreeNode root) {
dfs(root.left);
dfs(root.right);
visit(root);
}
非递归实现二叉树的前序遍历
144. Binary Tree Preorder Traversal (Medium)
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;
}
非递归实现二叉树的后序遍历
145. Binary Tree Postorder Traversal (Medium)
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。
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;
}
非递归实现二叉树的中序遍历
94. Binary Tree Inorder Traversal (Medium)
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。
修剪二叉查找树
669. Trim a Binary Search Tree (Easy)
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
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;
}
寻找二叉查找树的第 k 个元素
230. Kth Smallest Element in a BST (Medium)
中序遍历解法：
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：
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);
}
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)
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。
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);
}
二叉查找树的最近公共祖先
235. Lowest Common Ancestor of a Binary Search Tree (Easy)
_______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.
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;
}
二叉树的最近公共祖先
236. Lowest Common Ancestor of a Binary Tree (Medium)
_______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.
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;
}
从有序数组中构造二叉查找树
108. Convert Sorted Array to Binary Search Tree (Easy)
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;
}
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)
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
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;
}
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)
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。
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);
}
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)
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。
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);
}
寻找二叉查找树中出现次数最多的值
501. Find Mode in Binary Search Tree (Easy)
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。
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
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。
实现一个 Trie
208. Implement Trie (Prefix Tree) (Medium)
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';
}
}
实现一个 Trie，用来求前缀和
Input: insert("apple", 3), Output: Null
Input: sum("ap"), Output: 3
Input: insert("app", 2), Output: Null
Input: sum("ap"), Output: 5
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';
}
}
🎨
更多精彩内容将发布在公众号 CyC2018，公众号提供了该项目的离线阅读版本，后台回复"下载" 即可领取。也提供了一份面试复习思维导图，后台回复"资料" 即可领取。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。我基本是按照这个思维导图来进行复习的，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思维导图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。
更多精彩内容将发布在公众号 CyC2018，公众号提供了该项目的离线阅读版本，后台回复"下载" 即可领取。也提供了一份面试复习思维导图，后台回复"资料" 即可领取。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。我基本是按照这个思维导图来进行复习的，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思维导图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。
