Binary Tree Level Order Traversal（二叉树广度优先遍历或逐层遍历）

来源：https://leetcode.com/problems/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,null,null,15,7],

return its level order traversal as:

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

使用队列来存储每层的节点，然后每一层出队列后将其下一层入队列

Java

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 import java.util.LinkedList;
11 import java.util.Queue;
12 class Solution {
13     public List<List<Integer>> levelOrder(TreeNode root) {
14         Queue<TreeNode> queue = new LinkedList<TreeNode>();
15         List<List<Integer>> result = new ArrayList<List<Integer>>();
16         TreeNode node = null;
17         queue.offer(root);
18         int queueSize = 0;
19         while(!queue.isEmpty()) {
20             List<Integer> levelResult = new ArrayList<Integer>();
21             queueSize = queue.size();
22             while(queueSize-- > 0) {
23                 node = queue.poll();
24                 if(node == null) {
25                     continue;
26                 }
27                 levelResult.add(node.val);
28                 queue.add(node.left);
29                 queue.add(node.right);
30             }
31             if(!levelResult.isEmpty()) {
32                 result.add(levelResult);
33             }
34         }
35         return result;
36     }
37 }

Python

 1 # Definition for a binary tree node.
 2 # class TreeNode(object):
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 
 8 class Solution(object):
 9     def levelOrder(self, root):
10         """
11         :type root: TreeNode
12         :rtype: List[List[int]]
13         """
14         if root == None:
15             return []
16         level = [root]
17         results = []
18         while level:
19             results.append([node.val for node in level])
20             level = [kid for node in level for kid in (node.left, node.right) if kid]
21         return results