平凉网站建设平凉,自我介绍网页制作html,网站建设过程的结构图,黟县网站建设提示#xff1a;文章写完后#xff0c;目录可以自动生成#xff0c;如何生成可参考右边的帮助文档 文章目录 前言一、力扣270. 最接近的二叉搜索树值二、力扣404. 左叶子之和三、力扣617. 合并二叉树四、力扣623. 在二叉树中增加一行 前言 二叉树的递归分为「遍历」和「分解… 提示文章写完后目录可以自动生成如何生成可参考右边的帮助文档 文章目录 前言一、力扣270. 最接近的二叉搜索树值二、力扣404. 左叶子之和三、力扣617. 合并二叉树四、力扣623. 在二叉树中增加一行 前言 二叉树的递归分为「遍历」和「分解问题」两种思维模式这道题需要用到「遍历」的思维 一、力扣270. 最接近的二叉搜索树值 /*** 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 {double flag Integer.MAX_VALUE;double res 0;public int closestValue(TreeNode root, double target) {fun(root,target);return (int)res;}public void fun(TreeNode root , double target){if(root null){return;}fun(root.left,target);double temp Math.abs(target - root.val);if(temp flag){flag temp;res root.val;}fun(root.right,target);} }二、力扣404. 左叶子之和 /*** 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 int sumOfLeftLeaves(TreeNode root) {fun(root);return sum;}public void fun(TreeNode root){if(root null){return;}if(root.left ! null){if(root.left.left null root.left.right null){sum root.left.val;}}fun(root.left);fun(root.right);} }三、力扣617. 合并二叉树 /*** 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 mergeTrees(TreeNode root1, TreeNode root2) {return fun(root1,root2);}public TreeNode fun(TreeNode root1, TreeNode root2){if(root1 ! null root2 ! null){root1.val root1.val root2.val;}else if(root1 null root2 ! null){return root2;}else if(root1 ! null root2 null){return root1;}else{return null;}TreeNode l fun(root1.left, root2.left);TreeNode r fun(root1.right, root2.right);root1.left l;root1.right r;return root1;} }四、力扣623. 在二叉树中增加一行 /*** 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 len 0;public TreeNode addOneRow(TreeNode root, int val, int depth) {if(depth 1){return new TreeNode(val,root,null);}len depth-1;return fun(root, val, 1);}public TreeNode fun(TreeNode root, int val, int depth){if(root null){return null;}if(depth len){root.left new TreeNode(val,root.left,null);root.right new TreeNode(val,null,root.right);return root;}TreeNode l fun(root.left, val, depth1);TreeNode r fun(root.right, val, depth 1);root.left l;root.right r;return root;} }