【题解】【BT】【Leetcode】Binary Tree Level Order Traversal

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.

思路：

重点在于每层元素个数不定，如何标记一层的结束，往堆栈里push很多NULL来表示空位这种方案，会造成Memory Limit Exceeded。

可以采取记下每层的NULL数量，下层翻倍这种方式计数，满额push标记NULL作为一层的结束

代码：

 1 vector<vector<int> > levelOrder(TreeNode *root) {
 2     vector<vector<int> > orders;
 3     if(root == NULL)
 4         return orders;
 5         
 6     vector<int> vtmp;
 7     queue<TreeNode*> tque;
 8     tque.push(root);
 9     tque.push(NULL);
10     
11     int size = 2;
12     int count = 0;
13     int zero = 0;//该层的NULL数
14     while(!tque.empty()){
15         TreeNode * tmp = tque.front();//$$$$
16         tque.pop();//void pop()
17         
18         if(tmp == NULL){//NULL标识一层的结束
19             if(!vtmp.empty())//最后一行可能不满
20                 orders.push_back(vtmp);
21             vtmp.clear();
22             continue;
23         }
24         vtmp.push_back(tmp->val);
25         if(tmp->left != NULL){
26             tque.push(tmp->left);
27             count++;
28         }
29         else 
30             zero++;
31         if(tmp->right != NULL){
32             tque.push(tmp->right);
33             count++;
34         }
35         else
36             zero++;
37         
38         if(count + zero == size){
39             tque.push(NULL);
40             count = 0;
41             size *= 2; 
42             zero = zero * 2;
43         }
44     }
45     return orders;
46 }

 

posted on 2014-02-06 23:41  小唯THU  阅读(243)  评论(0)  编辑  收藏  举报

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