[LeetCode]Symmetric Tree

题目描述:(链接)

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.

解题思路:

递归版:

/**
* 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 == nullptr) {
return true;
}

return isSymmetric(root->left, root->right);
}

bool isSymmetric(TreeNode *left, TreeNode *right) {
if (!left && !right) return true;
if (!left || !right) return false;

return left->val == right->val &&
isSymmetric(left->left, right->right) &&
isSymmetric(left->right, right->left);
}
};

迭代版:

/**
* 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 == nullptr) return true;
stack<TreeNode *> cache;
cache.push(root->left);
cache.push(root->right);
while (!cache.empty()) {
auto p = cache.top(); cache.pop();
auto q = cache.top(); cache.pop();

if (!p && !q) continue;
if (!p || !q) return false;
if (p->val != q->val) return false;

cache.push(p->left); cache.push(q->right);
cache.push(p->right); cache.push(q->left);
}

return true;
}
};