Leetcode 102. Binary Tree Level Order Traversal

Posted 2020-07-07 琴影

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.

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:

   1
  / \\
 2   3
    /
   4
    \\
     5

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

 

分析：二叉树的层序遍历，使用广度优先搜索，建立一个队列q。
每次将队首temp结点出队列的同时，将temp的左孩子和右孩子入队
用size标记入队后的队列长度
row数组始终为当前层的遍历结果，然后用while(size–)语句保证每一次保存入row的只有一层。
每一次一层遍历完成后，将该row的结果存入二维数组v。

 1 /**
 2  * Definition for a binary tree node.
 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         vector<int> row;
14         vector<vector<int>> v;
15         queue<TreeNode *> q;
16         if(root == NULL)
17             return v;
18         q.push(root);
19         TreeNode *temp;
20         while(!q.empty()) {
21             int size = q.size();
22             while(size--) {
23                 temp = q.front();
24                 q.pop();
25                 row.push_back(temp->val);
26                 if(temp->left != NULL) {
27                     q.push(temp->left);
28                 }
29                 if(temp->right != NULL) {
30                     q.push(temp->right);
31                 }
32             }
33             v.push_back(row);
34             row.clear();
35         }
36         return v;
37     }
38 };

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