[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]
]

這道題目是二叉樹的層次遍歷。

思路：藉助於隊列結構(queue)，保存還沒有訪問兒子節點的節點。

這道題一大注意點在於，每層結束後插入NULL做爲分隔符。

1. 初始化：將根節點和第一層終止符NULL壓入隊列

2. 循環

用level記錄該層的節點val

- 取出隊首元素cur

- 若cur非空, 記錄cur->val

  若cur爲NULL，則說明該層節點已所有訪問過，將level插入ret. 

  若此時queue非空，訪問下一層節點，不然，中止循環

 

代碼

 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>> levelOrder(TreeNode *root) {
13         // IMPORTANT: Please reset any member data you declared, as
14         // the same Solution instance will be reused for each test case.
15         vector<vector<int>> ret;
16         if(root==NULL)
17             return ret;
18         
19         queue<TreeNode *> q;
20         q.push(root);
21         q.push(NULL); // IMPORTANT! end of level
22         
23         vector<int> level;
24         
25         while(!q.empty())
26         {
27             TreeNode *cur = q.front();
28             q.pop();
29             
30             if(cur)
31             {
32                 level.push_back(cur->val);
33                 if(cur->left)
34                     q.push(cur->left);
35                 if(cur->right)
36                     q.push(cur->right);
37             }
38             else
39             {
40                 ret.push_back(level);
41                 level.clear();
42                 if(q.empty())
43                     break;
44                 else
45                     q.push(NULL);
46             }
47         }
48         return ret;
49     }
50 };