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  • Symmetric Tree

    Symmetric Tree

    Total Accepted: 84678 Total Submissions: 259420 Difficulty: Easy

    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.

    1.递归版本

    /**
     * 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 {
    private:
        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->right,root2->left) && isSymmetric(root1->left,root2->right);
        }
    public:
        bool isSymmetric(TreeNode* root) {
            return isSymmetric(root,root);
        }
    };

     2.迭代版本,巧妙的利用入队列的顺序

    /**
     * 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;
            queue<TreeNode*> que;
            que.push(root->left);
            que.push(root->right);
            while(!que.empty()){
                TreeNode* p = que.front();que.pop();
                TreeNode* q = que.front();que.pop();
                if(!p && !q) continue;
                if(!p || !q) return false;
                if(p->val != q->val) return false;
                que.push(p->left);
                que.push(q->right);
                que.push(p->right);
                que.push(q->left);
            }
            return true;
        }
    };
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  • 原文地址:https://www.cnblogs.com/zengzy/p/5053343.html
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