LeetCode: Populating Next Right Pointers in Each Node II 解题报告

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * public class TreeLinkNode {
 4  *     int val;
 5  *     TreeLinkNode left, right, next;
 6  *     TreeLinkNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public void connect(TreeLinkNode root) {
11         if (root == null) {
12             return;
13         }
14         
15         TreeLinkNode dummy = new TreeLinkNode(0);
16         Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>();
17         
18         q.offer(root);
19         q.offer(dummy);
20         
21         while(!q.isEmpty()) {
22             TreeLinkNode cur = q.poll();
23             if (cur == dummy) {
24                 if (!q.isEmpty()) {
25                     q.offer(dummy);
26                 }
27                 continue;
28             }
29             
30             if (q.peek() == dummy) {
31                 cur.next = null;
32             } else {
33                 cur.next = q.peek();    
34             }
35             
36             if (cur.left != null) {
37                 q.offer(cur.left);
38             }
39             
40             if (cur.right != null) {
41                 q.offer(cur.right);
42             }
43         }
44         
45     }
46 }

 1 public void connect(TreeLinkNode root) {
 2         if (root == null) {
 3             return;
 4         }
 5         
 6         TreeLinkNode cur = root.next;
 7         TreeLinkNode next = null;
 8         // this is very important. should exit after found the next.
 9         while (cur != null && next == null) {
10             if (cur.left != null) {
11                 next = cur.left;
12             } else if (cur.right != null) {
13                 next = cur.right;
14             } else {
15                 cur = cur.next;
16             }
17         }
18         
19         if (root.right != null) {
20             root.right.next = next;
21             next = root.right;
22         }
23         
24         if (root.left != null) {
25             root.left.next = next;
26         }
27         
28         // The order is very important. We should deal with right first!
29         connect(root.right);
30         connect(root.left);
31     }