669. 修剪二叉搜索树
- 刷题
https://leetcode.cn/problems/trim-a-binary-search-tree/description/ - 文章讲解https://programmercarl.com/0669.%E4%BF%AE%E5%89%AA%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91.html
- 视频讲解https://www.bilibili.com/video/BV17P41177ud/?share_source=copy_web&vd_source=af4853e80f89e28094a5fe1e220d9062
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题解（递归法）：
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {//递归public TreeNode trimBST(TreeNode root, int low, int high) {//若传入二叉树为空，则直接返回nullif(root == null){return null;}//若当前结点小于规定区间，则当前结点的右分支还有可能在规定区间内//所以需要对当前结点右分支进行递归判断if(root.val < low){TreeNode right = trimBST(root.right, low, high);//将递归下一层得到的新的右分支根结点返回给当前层的父结点//当前root结点的删除return right;}//若当前结点大于规定区间，则当前结点的左分支还有可能在规定区间内//所以需要对当前结点左分支进行递归判断if(root.val > high){TreeNode left = trimBST(root.left, low,high);return left;}//若root在规定范围内//跟结点左分支递归root.left = trimBST(root.left, low, high);//根结点右分支递归root.right = trimBST(root.right, low, high);//返回新的二叉搜索树的根结点//前面也已经处理过只有根结点的情况，所以这里不用特殊处理根结点return root;}
}
108.将有序数组转换为二叉搜索树
- 刷题https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree/description/
- 文章讲解https://programmercarl.com/0108.%E5%B0%86%E6%9C%89%E5%BA%8F%E6%95%B0%E7%BB%84%E8%BD%AC%E6%8D%A2%E4%B8%BA%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91.html
- 视频讲解https://www.bilibili.com/video/BV1uR4y1X7qL/?share_source=copy_web&vd_source=af4853e80f89e28094a5fe1e220d9062
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题解（递归法）：
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {//递归法(左闭右闭)public TreeNode sortedArrayToBST(int[] nums) {return sortedArrayToBST(nums, 0, nums.length - 1);}public TreeNode sortedArrayToBST(int[] nums, int left, int right){//因为左闭右闭，所以left = right 为合法区间//则left > right为非法区间if(left > right){return null;}//若只有一个结点，则直接充当根结点，返回if(right - left == 0){return new TreeNode(nums[left]);}//若有两个及以上结点，进行递归//注意左闭右闭的递归对应int mid =( left + right ) / 2;TreeNode root = new TreeNode(nums[mid]);//左递归root.left = sortedArrayToBST(nums, left, mid - 1);//右递归root.right = sortedArrayToBST(nums, mid + 1, right);//返回根结点return root;}
}
538.把二叉搜索树转换为累加树
- 刷题https://leetcode.cn/problems/convert-bst-to-greater-tree/description/
- 文章讲解https://programmercarl.com/0538.%E6%8A%8A%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E8%BD%AC%E6%8D%A2%E4%B8%BA%E7%B4%AF%E5%8A%A0%E6%A0%91.html
- 视频讲解https://www.bilibili.com/video/BV1d44y1f7wP/?share_source=copy_web&vd_source=af4853e80f89e28094a5fe1e220d9062
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题解（递归法）：
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {//累加//记录累加和int sum;public TreeNode convertBST(TreeNode root) {sum = 0;convertBST1(root);return root;}//由题意得，按右中左顺序遍历，进行累加即可public void convertBST1(TreeNode root){//若传入空树，则直接返回if(root == null){return;}//右处理convertBST1(root.right);//中处理//记录当前累加和sum += root.val;//将当前累加和赋值给当前结点root.val = sum;//左处理convertBST1(root.left);}
}
