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