沈阳市网站制作,艺术字体logo设计生成器,舞台灯光网站建设公司,seo策划方案文章目录 Leetcode 669. 修剪二叉搜索树解题思路代码总结 Leetcode 108. 将有序数组转换为二叉搜索树解题思路代码总结 Leetcode 538. 把二叉搜索树转换为累加树解题思路代码总结 草稿图网站 java的Deque
Leetcode 669. 修剪二叉搜索树 题目#xff1a;669. 修剪二叉搜索树 解… 文章目录 Leetcode 669. 修剪二叉搜索树解题思路代码总结 Leetcode 108. 将有序数组转换为二叉搜索树解题思路代码总结 Leetcode 538. 把二叉搜索树转换为累加树解题思路代码总结 草稿图网站 java的Deque
Leetcode 669. 修剪二叉搜索树 题目669. 修剪二叉搜索树 解析代码随想录解析 解题思路
对于不符合的节点如果该节点小于区间则右孩子可能符合如果该节点大于区间则左孩子可能符合。
代码
/*** 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 {public TreeNode trimBST(TreeNode root, int low, int high) {if (root null)return root;if (root.val low) {return trimBST(root.right, low, high);}if (root.val high) {return trimBST(root.left, low, high);}root.left trimBST(root.left, low, high);root.right trimBST(root.right, low, high);return root;}
}//迭代法
class Solution {public TreeNode trimBST(TreeNode root, int low, int high) {if (root null)return root;while (root ! null (root.val low || root.val high)) {if (root.val low)root root.right;elseroot root.left;}TreeNode cur root;while (cur ! null) {while (cur.left ! null cur.left.val low)cur.left cur.left.right;cur cur.left;}cur root;while (cur ! null) {while (cur.right ! null cur.right.val high)cur.right cur.right.left;cur cur.right;}return root;}
}总结
暂无
Leetcode 108. 将有序数组转换为二叉搜索树 题目108. 将有序数组转换为二叉搜索树 解析代码随想录解析 解题思路
递归数组
代码
/*** 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 {public TreeNode sortedArrayToBST(int[] nums) {if (nums null || nums.length 0)return null;return buildTree(nums, 0, nums.length);}private TreeNode buildTree(int[] nums, int left, int right) {if (left right)return null;if (left 1 right)return new TreeNode(nums[left]);int mid left (right - left) / 2;TreeNode midNode new TreeNode(nums[mid]);midNode.left buildTree(nums, left, mid);midNode.right buildTree(nums, mid 1, right);return midNode;}
}总结
迭代懒得写了
Leetcode 538. 把二叉搜索树转换为累加树 题目538. 把二叉搜索树转换为累加树 解析代码随想录解析 解题思路
反过来的中序
代码
/*** 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 {int sum 0;public TreeNode convertBST(TreeNode root) {if (root null)return null;order(root);return root;}private void order(TreeNode node) {if (node null)return;order(node.right);sum node.val;node.val sum;order(node.left);}
}总结
递归懒得写
