## [101\. Symmetric Tree](https://leetcode.com/problems/symmetric-tree/)

Given the `root` of a binary tree, _check whether it is a mirror of itself_ (i.e., symmetric around its center).

**Example 1:**


**Input:** root = \[1,2,2,3,4,4,3\]
**Output:** true

**Example 2:**


**Input:** root = \[1,2,2,null,3,null,3\]
**Output:** false

**Constraints:**

- The number of nodes in the tree is in the range `[1, 1000]`.
- `-100 <= Node.val <= 100`

解法一：recursion

/**
 * 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) {
        return isSame(root, root);
    }
    bool isSame(TreeNode* t1, TreeNode* t2) {
        if (!t1 && !t2) return true;
        if (!t1 || !t2) return false;
        if (t1->val != t2->val) return false;
        return isSame(t1->left, t2->right) && isSame(t1->right, t2->left);
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::

解法二：iteration，用一個 queue，或兩個 queue

/**
 * 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) {
        queue<TreeNode*> q;
        q.push(root);
        q.push(root);
        while (!q.empty()) {
            TreeNode* t1 = q.front(); q.pop();
            TreeNode* t2 = q.front(); q.pop();
            if (!t1 && !t2) continue;
            if (!t1 || !t2) return false;
            if (t1->val != t2->val) return false;
            q.push(t1->left);
            q.push(t2->right);
            q.push(t1->right);
            q.push(t2->left);
        }
        return true;        
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::