101. Symmetric Tree

[huimeich]

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

For example, this binary tree is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3
But the following is not:
    1
   / \
  2   2
   \   \
   3    3

[huimeich]

/**
 * 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) {
        vector<TreeNode*> l;
        vector<TreeNode*> r;
        TreeNode* lr; TreeNode* rr;
        if (root == NULL) return true;
        l.push_back(root);
        r.push_back(root);
        while (!l.empty() && !r.empty()) {
            lr=l.back();   rr=r.back();
            l.pop_back(); r.pop_back();
            if (lr->val != rr->val) return false;
            if (lr->left != NULL && rr->right != NULL) {
                l.push_back(lr->left);
                r.push_back(rr->right);
            }
            else if (lr->left != NULL || rr->right != NULL) return false;
            if (lr->right != NULL && rr->left != NULL) {
                l.push_back(lr->right);
                r.push_back(rr->left);
            }
            else if (lr->right != NULL || rr->left != NULL) return false;
        }
        if (l.empty() && r.empty()) return true;
        return false;
    }
};