[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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