[leetcode][tree] Symmetric Tree

题目：

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.
递归实现

/**
* 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(NULL == root) return true;
return isSymetricCore(root->left, root->right);
}
private:
bool isSymetricCore(TreeNode *left, TreeNode *right){
if(NULL == left && NULL == right) return true;
if(NULL == left || NULL == right || left->val != right->val) return false;
return isSymetricCore(left->left, right->right) && isSymetricCore(left->right, right->left);
}
};

非递归实现（方法类似于中序遍历）

/**
* 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 (NULL == root || NULL == root->left && NULL == root->right) return true;
if (NULL == root->left || NULL == root->right) return false;
stack<TreeNode *> leftStack, rightStack;
leftStack.push(root->left);
rightStack.push(root->right);
TreeNode *pLeft = root->left;
TreeNode *pRight = root->right;
while (pLeft && pRight || !leftStack.empty() && !rightStack.empty()){
while (pLeft && pRight){
leftStack.push(pLeft);
rightStack.push(pRight);
pLeft = pLeft->left;
pRight = pRight->right;
}
if (pLeft || pRight) return false;
pLeft = leftStack.top();
pRight = rightStack.top();
leftStack.pop();
rightStack.pop();
if (pLeft->val != pRight->val) return false;
pLeft = pLeft->right;
pRight = pRight->left;
}
if (pLeft || pRight || !leftStack.empty() || !rightStack.empty()) return false;
return true;
}
};