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],

    3
   / \
  9  20
    /  \
   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

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 import java.util.LinkedList;
 import java.util.Queue;
 class Solution {
     public List<List<Integer>> levelOrder(TreeNode root) {
         Queue<TreeNode> queue = new LinkedList<TreeNode>();
         List<List<Integer>> result = new ArrayList<List<Integer>>();
         TreeNode node = null;
         queue.offer(root);
         int queueSize = 0;
         while(!queue.isEmpty()) {
             List<Integer> levelResult = new ArrayList<Integer>();
             queueSize = queue.size();
             while(queueSize-- > 0) {
                 node = queue.poll();
                 if(node == null) {
                     continue;
                 }
                 levelResult.add(node.val);
                 queue.add(node.left);
                 queue.add(node.right);
             }
             if(!levelResult.isEmpty()) {
                 result.add(levelResult);
             }
         }
         return result;
     }
 }

Python

 # Definition for a binary tree node.
 # class TreeNode(object):
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None

 class Solution(object):
     def levelOrder(self, root):
         """
         :type root: TreeNode
         :rtype: List[List[int]]
         """
         if root == None:
             return []
         level = [root]
         results = []
         while level:
             results.append([node.val for node in level])
             level = [kid for node in level for kid in (node.left, node.right) if kid]
         return results