/*
Author: Andy, [email protected]
Date: Dec 12, 2014
Problem: Binary Tree Level Order Traversal
Difficulty: easy
Source: https://oj.leetcode.com/problems/binary-tree-level-order-traversal/
Notes:
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]
]
Solution: 1. Use queue. In order to seperate the levels, use 'NULL' as the end indicator of one level.
2. DFS.
*/
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List> levelOrder_1(TreeNode root) {
List
> res = new ArrayList
>();
if (root == null) return res;
Queue
> levelOrder_2(TreeNode root) {
List
> res = new ArrayList
>();
if (root == null) return res;
levelOrderRe(root, 0, res);
return res;
}
public void levelOrderRe(TreeNode root, int level, List
> res) {
if(root == null) return;
if(level == res.size()) res.add(new ArrayList