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  • LeetCode 101. Symmetric Tree(镜像树)

    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.

    /**
     * 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) return true; 
            
            TreeNode *left;
            TreeNode *right;
            
            queue<TreeNode *> q1, q2;
            q1.push(root->left); q2.push(root->right);
            
            while (!q1.empty() && !q2.empty())
            {
                left = q1.front(); q1.pop();
                right = q2.front(); q2.pop();
                
                if (NULL == left && NULL == right)
                    continue;
                if (NULL == left || NULL == right)
                    return false;
                if (left->val != right->val)
                    return false;
                
                q1.push(left->left); 
                q1.push(left->right);
                q2.push(right->right); 
                q2.push(right->left);
            }
            
            return true;
        }
    };
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  • 原文地址:https://www.cnblogs.com/douzujun/p/10980129.html
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