# Leetcode [No. 101] Symmetric Tree (EASY)

+ First Solved on 2023/12/29
+ second solved on 2025/01/04

## 題目

https://leetcode.com/problems/symmetric-tree/

## 思路
+ 這個解法其實不太算是我想的，但我覺得這個寫法很讚
+ 最主要就是在dfs裡面去判斷兩個node是否相同，如果相同的話就繼續往下比....
+ 跟一般的DFS差在他用兩個ptr去指

/**
 * 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 isSymmetric(TreeNode* root) {
        // compare left to right
        return isSameTree(root->left, root->right);
    }
    bool isSameTree(TreeNode* p, TreeNode* q)
    {
        if(p==q)
        {
            return true;
        }
        else if (p==nullptr && q!= nullptr || p != nullptr && q == nullptr)
        {
            return false;
        }
        else
        {
            return (p->val == q->val) && isSameTree(p->left, q->right) && isSameTree(p->right, q->left);
        }
    }
};

### 解法分析
+ time complexity: O(lgn)

### 執行結果

# 二次挑戰

## Intution

+ 根據代碼啟示錄的教學，再進行二度改良。

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        return compare(root->left, root->right);
    }
    bool compare(TreeNode*left, TreeNode* right) 
    {
        if(left == right) {
            return true;
        } else if (left && right) {
            return left->val == right->val && compare(left->right, right->left) && compare(left->left, right->right);
        } else {
            return false;
        }
    }
};

## Analyze

+ Time complexity: O(n)

## Performance


## follow up:

> Follow up: Could you solve it both recursively and iteratively?

### 解法分析
+ time complexity: O(nlgn)

### 執行結果