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notes/Leetcode 题解.md
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@ -5622,116 +5622,6 @@ private boolean isLeaf(TreeNode node){
}
```
**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
```
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。
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;
}
```
**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;
}
```
**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;
}
```
**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;
}
```
**相同节点值的最大路径长度**
[687. Longest Univalue Path (Easy)](https://leetcode.com/problems/longest-univalue-path/)
@ -5998,190 +5888,54 @@ public List<Integer> inorderTraversal(TreeNode root) {
### BST
主要利用 BST 中序遍历有序的特点。
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。
**在 BST 中寻找两个节点，使它们的和为一个给定值**
二叉查找树中序遍历有序。
[653. Two Sum IV - Input is a BST (Easy)](https://leetcode.com/problems/two-sum-iv-input-is-a-bst/description/)
**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:
5
3
/ \
3 6
/ \ \
2 4 7
0 4
\
2
/
1
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);
}
```
**在 BST 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
L = 1
R = 3
Output:
1
3
/
2
/
1
```
利用 BST 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。
只保留值在 L \~ R 之间的节点
```java
private int minDiff = Integer.MAX_VALUE;
private int preVal = -1;
public int getMinimumDifference(TreeNode root) {
inorder(root);
return minDiff;
}
private void inorder(TreeNode node){
if (node == null) return;
inorder(node.left);
if (preVal != -1) minDiff = Math.min(minDiff, Math.abs(node.val - preVal));
preVal = node.val;
inorder(node.right);
}
```
**把 BST 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);
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;
}
private void traver(TreeNode root) {
if (root == null) return;
if (root.right != null) traver(root.right);
sum += root.val;
root.val = sum;
if (root.left != null) traver(root.left);
}
```
**寻找 BST 中出现次数最多的节点**
[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].
```
```java
private int cnt = 1;
private int maxCnt = 1;
private TreeNode preNode = null;
private List<Integer> list;
public int[] findMode(TreeNode root) {
list = new ArrayList<>();
inOrder(root);
int[] ret = new int[list.size()];
int idx = 0;
for (int num : list) {
ret[idx++] = num;
}
return ret;
}
private void inOrder(TreeNode node) {
if (node == null) return;
inOrder(node.left);
if (preNode != null) {
if (preNode.val == node.val) cnt++;
else cnt = 1;
}
if (cnt > maxCnt) {
maxCnt = cnt;
list.clear();
list.add(node.val);
} else if (cnt == maxCnt) {
list.add(node.val);
}
preNode = node;
inOrder(node.right);
}
```
**寻找 BST 的第 k 个元素**
**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/)
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);
}
```
中序遍历解法：
@ -6206,7 +5960,129 @@ private void inOrder(TreeNode node, int k) {
}
```
**根据有序链表构造平衡的 BST**
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);
}
```
**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 root) {
if (root == null) return;
if (root.right != null) traver(root.right);
sum += root.val;
root.val = sum;
if (root.left != null) traver(root.left);
}
```
**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;
}
```
**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;
}
```
**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;
}
```
**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/)
@ -6251,6 +6127,138 @@ private int size(ListNode node) {
}
```
**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);
}
```
**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, Math.abs(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)](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，并且是最大的。
```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