5.1.7 Symmetric Tree

ifficulty: Easy Source: http://leetcode.com/onlinejudge#question_101 Notes: 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 Note: Bonus points if you could solve it both recursively and iteratively.  Solution: 1. Recursive solution 2.Iterative way (queue). */ /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */

class Solution {public:    bool isSymmetric(TreeNode *root) {        return isSymmetric_2(root);    }    bool isSymmetric_1(TreeNode *root) {        if (root == NULL) return true;        return solve (root->left, root->right);    }    bool solve(TreeNode * t1, TreeNode * t2) {        if (!t1 && !t2) return true;        if (!t1 && t2 || t1 && !t2 || t1->val != t2->val) return false;        return solve(t1->left, t2->right) && solve(t1->right, t2->left);    }

};