101. Symmetric Tree

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.

宽度遍历

class Solution {public:    bool isSymmetric(TreeNode* root) {        queue<TreeNode*> left,right;        if(!root||(!root->left && !root->right) ) return true;        TreeNode* lp;TreeNode* rp;        if(root->left) left.push(root->left);        if(root->right) right.push(root->right);        while(!left.empty()&&!right.empty())        {            lp=left.front();            rp=right.front();            if(lp->val!=rp->val)                return false;            if(lp->left) left.push(lp->left);            if(lp->right) left.push(lp->right);            if(rp->right) right.push(rp->right);            if(rp->left) right.push(rp->left);            left.pop();            right.pop();        }        if(left.size()!=right.size())            return false;        return true;    }};

此代码存在问题！

   1   / \  2   2   \   \   3    3

对于上面的test就无法得出正确答案。

正确代码如下：

class Solution {public:    bool isSymmetric(TreeNode* root) {        if (!root) return true;        queue<TreeNode*> check;                check.push(root->left);        check.push(root->right);                while (!check.empty()) {            TreeNode* node1 = check.front();            check.pop();            TreeNode* node2 = check.front();            check.pop();            if (!node1 && node2) return false;            if (!node2 && node1) return false;            if (node1 && node2) {                if (node1->val != node2->val) return false;                check.push(node1->left);                check.push(node2->right);                check.push(node1->right);                check.push(node2->left);            }        }        return true;                    }};

树的遍历的基础。