二叉树的锯齿形层序遍历
题目介绍
给定一个二叉树，返回其节点值的锯齿形层序遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。
例如：
给定二叉树 [3,9,20,null,null,15,7],
返回锯齿形层序遍历如下：
题目解法
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
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
| package algorithm;
import java.util.*;
public class BinaryTreeZigzagLevelOrderTraversal {
public static List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> ans = new LinkedList<>();
if (root == null) {
return ans;
}
Queue<TreeNode> nodeQueue = new LinkedList<>();
nodeQueue.offer(root);
boolean isOrderLeft = true;
while (!nodeQueue.isEmpty()) {
Deque<Integer> levelList = new LinkedList<>();
int size = nodeQueue.size();
for (int i = 0; i < size; ++i) {
TreeNode curNode = nodeQueue.poll();
if (isOrderLeft) {
levelList.offerLast(curNode.val);
} else {
levelList.offerFirst(curNode.val);
}
if (curNode.left != null) {
nodeQueue.offer(curNode.left);
}
if (curNode.right != null) {
nodeQueue.offer(curNode.right);
}
}
ans.add(new LinkedList<>(levelList));
isOrderLeft = !isOrderLeft;
}
return ans;
}
public static void main(String[] args) {
TreeNode n1 = new TreeNode(1);
TreeNode n2 = new TreeNode(2);
n1.left = n2;
TreeNode n3 = new TreeNode(3);
n1.right = n3;
TreeNode n4 = new TreeNode(4);
n2.left = n4;
n2.right = null;
TreeNode n5 = new TreeNode(5);
n3.right = n5;
n3.left = null;
List<List<Integer>> lists = zigzagLevelOrder(n1);
for (List<Integer> list : lists) {
for (Integer integer : list) {
System.out.print(integer + " ");
}
System.out.println();
}
}
public static 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;
}
}
}
|
打印：
思路：
思路上，利用了java的linkedqueue的队列特性，而且是两个队列。