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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

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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, depth+1);TreeNode r = fun(root.right, val, depth +1);root.left = l;root.right = r;return root;}
}