forked from andavid/leetcode-java
-
Notifications
You must be signed in to change notification settings - Fork 0
/
116.populating-next-right-pointers-in-each-node.java
110 lines (104 loc) · 3.23 KB
/
116.populating-next-right-pointers-in-each-node.java
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
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
/*
* @lc app=leetcode id=116 lang=java
*
* [116] Populating Next Right Pointers in Each Node
*
* https://leetcode.com/problems/populating-next-right-pointers-in-each-node/description/
*
* algorithms
* Medium (39.05%)
* Likes: 1144
* Dislikes: 97
* Total Accepted: 263.6K
* Total Submissions: 675.2K
* Testcase Example: '{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}'
*
* You are given a perfect binary tree where all leaves are on the same level,
* and every parent has two children. The binary tree has the following
* definition:
*
*
* struct Node {
* int val;
* Node *left;
* Node *right;
* Node *next;
* }
*
*
* Populate each next pointer to point to its next right node. If there is no
* next right node, the next pointer should be set to NULL.
*
* Initially, all next pointers are set to NULL.
*
*
*
* Example:
*
*
*
*
* Input:
* {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
*
* Output:
* {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}
*
* Explanation: Given the above perfect binary tree (Figure A), your function
* should populate each next pointer to point to its next right node, just like
* in Figure B.
*
*
*
*
* Note:
*
*
* You may only use constant extra space.
* Recursive approach is fine, implicit stack space does not count as extra
* space for this problem.
*
*
*/
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val,Node _left,Node _right,Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
if (root == null) return null;
Queue<Node> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
Node prev = null;
for (int i = 0; i < size; i++) {
Node node = queue.poll();
if (prev != null) {
prev.next = node;
}
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
prev = node;
}
prev.next = null;
}
return root;
}
}