LeetCode

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1   / \  2   2 / \ / \3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1   / \  2   2   \   \   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

判断一个二叉树是否是对称的。

不想写递归，于是依旧写循环。用两个队列来模拟，将左子树（相对于root）push进第一个队列，右子树push进第二个队列。然后每次判断两个队列弹出的值是否相等。时间复杂度O(节点数)，空间复杂度O(节点数)

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isSymmetric(TreeNode* root) {        if (!root) return true;        queue<TreeNode*> q1;        queue<TreeNode*> q2;        q1.push(root);        q2.push(root);        while (!q1.empty() && !q2.empty()) {            TreeNode* cur1 = q1.front();            TreeNode* cur2 = q2.front();            q1.pop();            q2.pop();            if (!cur1 && !cur2) continue;            else if (!cur1 || !cur2) return false;            if (cur1->val != cur2->val) return false;            q1.push(cur1->left);            q1.push(cur1->right);            q2.push(cur2->right);            q2.push(cur2->left);        }        return true;    }};