From 41d8d6a8602b8dbaf8cf746e828750a5bde6aaf1 Mon Sep 17 00:00:00 2001
From: gameloader <ggwqqo@163.com>
Date: Thu, 24 Oct 2024 13:53:56 +0800
Subject: [PATCH] leetcode update

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 content/posts/leetcode.md | 46 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 46 insertions(+)

diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index 0568df4..18c23e0 100644
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@@ -16340,3 +16340,49 @@ public:
     }
 };
 ```
+
+## day230 2024-10-24
+
+### 951. Flip Equivalent Binary Trees
+
+For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.
+
+A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.
+
+Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.
+
+![1024M2CLc5WxLfOZ](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/1024M2CLc5WxLfOZ.png)
+
+### 题解
+
+本题在使用dfs遍历树的基础上对dfs进行一些扩充，考虑如果两棵树是翻转等价的，则对于任何一个父节点，其两个子节点的值是一样的，只是值对应的左右子节点可能位置不同（一棵树是值1在左子节点，另一棵则是值1在右子节点），而在使用dfs遍历时无需考虑除当前节点左右子节点之外的节点，其余节点在递归时总会遍历到。
+
+在dfs时，将两棵树当前节点的指针作为参数，只需判断当前两节点是否相等（包括都为空，或者一个为空一个不为空的情况），相等继续调用dfs判断左右子树是否相同或者为镜像翻转即可。即判断左树和左树相同，右树和右树相同（左右子树相同）或者左树和右树相同，右树和左树相同（左右子树翻转）即可。不相等直接返回false。
+
+### 代码
+
+```cpp
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    bool flipEquiv(TreeNode* root1, TreeNode* root2) {
+        if (root1 == nullptr && root2 == nullptr){
+            return true;
+        }
+        if((root1 != nullptr && root2 == nullptr)||(root1==nullptr && root2 != nullptr)||(root1->val != root2->val)){
+            return false;
+        }
+        return (flipEquiv(root1->left, root2->left) && flipEquiv(root1->right,root2->right)) || (flipEquiv(root1->right, root2->left) && flipEquiv(root1->left,root2->right));
+    }
+};
+```