LeetCode: Symmetric Tree

Title:

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

/**
 * 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 fun(root->left,root->right);
        
    }
    bool fun(TreeNode* p1, TreeNode* p2){
        if (!p1 && !p2 )
            return true;
        if (!p1 || !p2 )
            return false;
        return (p1->val == p2->val) && fun(p1->left,p2->right) && fun(p1->right,p2->left);
        
    }
};
/**
 * 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;
        stack<TreeNode* > s;
        s.push(root->left);
        s.push(root->right);
        while (!s.empty()){
            TreeNode* p1 = s.top();
            s.pop();
            TreeNode* p2 = s.top();
            s.pop();
            if (!p1 && !p2)
                continue;
            if (!p1 || !p2)
                return false;
            if (p1->val != p2->val)
                return false;
            s.push(p1->left);
            s.push(p2->right);
            s.push(p1->right);
            s.push(p2->left);
        }
        return true;
    }
    
};

 

posted on 2015-05-12 14:46  月下之风  阅读(130)  评论(0编辑  收藏  举报

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