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.

判断二叉树是否为平衡二叉树。

递归实现。

代码如下:

 /**
* 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) {
if(root == NULL)
{
return true;
}
return isSame(root->left, root->right);
}
bool isSame(TreeNode* left, TreeNode* right)
{
if(left == NULL && right == NULL)
{
return true;
}
else if(left == NULL)
{
return false;
}
else if(right == NULL)
{
return false;
} if(left->val == right->val && left->left == NULL && left->right && right->right == NULL && right->left == NULL)
{
return true;
}
else if(left->val == right->val)
{
return isSame(left->left, right->right) && isSame(left->right, right->left);
}
else
{
return false;
} }
};

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