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Binary Tree Level Order Traversal <leetcode>

Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3   /   9  20    /     15   7

 

return its level order traversal as:

[  [3],  [9,20],  [15,7]]

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

算法:本来层次用一个链表就可以了,很容易解决,但是这里的返回结果要求记录每个节点的深度(层),所以用了其他的方法,在网上也查过其他的方法,两个链表也可以解决,这里没有代码,我的代码如下:

 

 1 /** 2  * Definition for binary tree 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     vector<vector<int>>  result;13     vector<vector<int> > levelOrder(TreeNode *root) {14         result.clear();15         doit(root,1);16         return result;17     }18     19     void doit(TreeNode *root,int level)20     {21         if(NULL==root)22         {23             return;24         }25         else26         {27             if(level>result.size())28             {29                 vector<int>  node;30                 result.push_back(node);31             }32             result[level-1].push_back(root->val);33             doit(root->left,level+1);34             doit(root->right,level+1);35         }36     }37 };

 

Binary Tree Level Order Traversal <leetcode>