LeetCode --- 102. 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},

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.

OJ‘s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.

Here‘s an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

这道题的要求是分层遍历二叉树。

由于需要把每层的节点分别放入到数组中，因此需要引入变量n记录每层的节点数量。剩下的，就是广度优先搜索的方法了。

广度优先搜索算法（Breadth First Search），又叫宽度优先搜索，或横向优先搜索。从根节点开始，沿着树的宽度遍历树的节点。如果所有节点均被访问，则算法中止。借助队列数据结构，由于队列是先进先出的顺序，因此可以先将左子树入队，然后再将右子树入队。这样一来，左子树结点就存在队头，可以先被访问到。

时间复杂度：O(n)

空间复杂度：O(n)

 1 class Solution {
 2 public:
 3     vector<vector<int> > levelOrder(TreeNode *root)
 4     {
 5         vector<vector<int> > vvi;
 6         
 7         if(NULL == root)
 8             return vvi;
 9         
10         queue<TreeNode *> q;
11         q.push(root);
12         while(!q.empty())
13         {
14             vector<int> vi;
15             for(int i = 0, n = q.size(); i < n; ++ i)
16             {
17                 TreeNode *temp = q.front();
18                 q.pop();
19                 if(temp -> left != NULL)
20                     q.push(temp -> left);
21                 if(temp -> right != NULL)
22                     q.push(temp -> right);
23                 vi.push_back(temp -> val);
24             }
25             vvi.push_back(vi);
26         }
27         
28         return vvi;
29     }
30 };