103. Binary Tree Zigzag Level Order Traversal 103. Binary Tree Zigzag Level Order Traversal - Medium
Description 103. Binary Tree Zigzag Level Order Traversal
Given the root of a binary tree, return the zigzag level order traversal of its nodes' values. (i.e., from left to right, then right to left for the next level and alternate between).
Example 1 :
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]
还是一样的思路，层序遍历，加一个变量来控制方向就行。
Solution 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41 /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/ class Solution { public List < List < Integer >> zigzagLevelOrder ( TreeNode root ) { if ( root == null ) return new ArrayList <> (); Deque < TreeNode > q = new LinkedList <> (); List < List < Integer >> ans = new ArrayList <> (); q . offer ( root ); boolean f = true ; while ( ! q . isEmpty ()) { int size = q . size (); LinkedList < Integer > list = new LinkedList (); for ( int i = 0 ; i < size ; i ++ ) { TreeNode node = q . poll (); if ( f ) { list . addLast ( node . val ); } else { list . addFirst ( node . val ); } if ( node . left != null ) q . offer ( node . left ); if ( node . right != null ) q . offer ( node . right ); } f = ! f ; ans . add ( list ); } return ans ; } }
