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前言

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有些题目，你按照拍脑袋的方式去做，可能发现需要在递归代码中调用其他递归函数计算字数的信息。一般来说，出现这种情况时你可以考虑用后序遍历的思维方式来优化算法，利用后序遍历传递子树的信息，避免过高的时间复杂度

一、力扣1379. 找出克隆二叉树中的相同节点

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/class Solution {TreeNode res = null;public final TreeNode getTargetCopy(final TreeNode original, final TreeNode cloned, final TreeNode target) {fun(original,cloned,target);return res;}public void fun(TreeNode original, TreeNode cloned, TreeNode target){if(original == null){return;}if(original == target){res = cloned;return;}fun(original.left,cloned.left,target);fun(original.right,cloned.right,target);}
}

二、力扣LCR 143. 子结构判断

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public boolean isSubStructure(TreeNode A, TreeNode B) {if(B == null){return false;}if(A == null){return B == null;}if(fun(A,B)){return true;}return isSubStructure(A.left,B) || isSubStructure(A.right,B);}public boolean fun(TreeNode A , TreeNode B){if (B == null) {return true;}if (B != null && A == null) {return false;}if(A.val != B.val){return false;}return fun(A.left,B.left) && fun(A.right,B.right);}
}

三、力扣110. 平衡二叉树

/*** 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 {boolean flag = true;public boolean isBalanced(TreeNode root) {fun(root);return flag;}public int fun(TreeNode root){if(root == null){return 0;}int l = fun(root.left);int r = fun(root.right);if(Math.abs(l-r) > 1){flag = false;}return l > r ? l + 1:r +1;}
}

四、力扣250. 统计同值子树

/*** 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 int countUnivalSubtrees(TreeNode root) {if(root == null){return 0;}int cur = 0;if(fun(root,root.val)){cur = 1;}int l = countUnivalSubtrees(root.left);int r = countUnivalSubtrees(root.right);return cur + l + r;}public boolean fun(TreeNode root,int val){if(root == null){return true;}if(root.val != val){return false;}return fun(root.left,val) && fun(root.right,val);}
}