[leetCode] Symmetric Tree

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.

思路一：递归思想。时间复杂度O(n)，空间复杂度O(logN)

　　

/**
* 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 *root) {
if (root == NULL) return true;
return isSymmetric(root->left, root->right);
}
bool isSymmetric(TreeNode *root1, TreeNode *root2) {
if (!root1 && !root2) return true;
if (!root1 || !root2) return false;
if (root1->val != root2->val) return false;

return isSymmetric(root1->left, root2->right) && isSymmetric(root1->right, root2->left);
}
};

思路二：迭代。层次遍历的思想。时间复杂度O(n),空间复杂度O(logN）

　　

/**
* 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 *root) {
if (root == NULL) return true;
queue<TreeNode *> q;
q.push(root->left);
q.push(root->right);

while (!q.empty()) {
TreeNode *p1 = q.front();
q.pop();
TreeNode *p2 = q.front();
q.pop();

if (!p1 && !p2) continue;
if (!p1 || !p2) return false;
if (p1->val != p2->val) return false;

q.push(p1->left);
q.push(p2->right);

q.push(p1->right);
q.push(p2->left);
}

return true;
}

};