[LeetCode] 116: Symmetric Tree

[Problem]

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

[Solution]

/**
 * 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 *root1, TreeNode *root2){
if(NULL == root1 && NULL == root2)return true;
if(NULL == root1 || NULL == root2)return false;
return root1->val == root2->val && isSymmetric(root1->left, root2->right) && isSymmetric(root1->right, root2->left);
}
bool isSymmetric(TreeNode *root) {
// Note: The Solution object is instantiated only once and is reused by each test case.
return root == NULL || isSymmetric(root->left, root->right);
}
};

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