117. Populating Next Right Pointers in Each Node II

题目：

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.

For example,
Given the following binary tree,

1
/  \
2    3
/ \    \
4   5    7

After calling your function, the tree should look like:

1 -> NULL
/  \
2 -> 3 -> NULL
/ \    \
4-> 5 -> 7 -> NULL

链接： http://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

4/16/2017

BB电面准备

抄别人的答案，但是递归应该不是constant extra space

注意的问题

1. 第13-15要在root层一直找到root.next整个表中子树不为0的情况

2. 第17行找到子树层可以用于子树层next的节点

3. 第18，19行给子树层找到next

4. 最重要的是先找right再找left

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) return;
12         TreeLinkNode nextNode = root.next;
13         while (nextNode != null) {
14             if (nextNode.left == null && nextNode.right == null) nextNode = nextNode.next;
15             else break;
16         }
17         if (nextNode != null) nextNode = (nextNode.left == null? nextNode.right: nextNode.left);
18         if (root.right != null) root.right.next = nextNode;
19         if (root.left != null) root.left.next = (root.right == null? nextNode: root.right);
20         connect(root.right);
21         connect(root.left);
22     }
23 }

别人的答案，iterative方法

这种还是需要分层的方法，什么手段可以分层？可以多用几个指针，并且在next层每一个元素都同时有current层的元素指到。

https://discuss.leetcode.com/topic/8447/simple-solution-using-constant-space

1 public class Solution {
2     public void connect(TreeLinkNode root) {
3
4         while(root != null){
5             TreeLinkNode tempChild = new TreeLinkNode(0);
6             TreeLinkNode currentChild = tempChild;
7             while(root!=null){
8                 if(root.left != null) { currentChild.next = root.left; currentChild = currentChild.next;}
9                 if(root.right != null) { currentChild.next = root.right; currentChild = currentChild.next;}
10                 root = root.next;
11             }
12             root = tempChild.next;
13         }
14     }
15 }

https://discuss.leetcode.com/topic/1106/o-1-space-o-n-complexity-iterative-solution

1 public class Solution {
2
3     //based on level order traversal
4     public void connect(TreeLinkNode root) {
5
6         TreeLinkNode head = null; //head of the next level
7         TreeLinkNode prev = null; //the leading node on the next level
8         TreeLinkNode cur = root;  //current node of current level
9
10         while (cur != null) {
11
12             while (cur != null) { //iterate on the current level
13                 //left child
14                 if (cur.left != null) {
15                     if (prev != null) {
16                         prev.next = cur.left;
17                     } else {
18                         head = cur.left;
19                     }
20                     prev = cur.left;
21                 }
22                 //right child
23                 if (cur.right != null) {
24                     if (prev != null) {
25                         prev.next = cur.right;
26                     } else {
27                         head = cur.right;
28                     }
29                     prev = cur.right;
30                 }
31                 //move to next node
32                 cur = cur.next;
33             }
34
35             //move to next level
36             cur = head;
37             head = null;
38             prev = null;
39         }
40
41     }
42 }

更多讨论：

https://discuss.leetcode.com/category/125/populating-next-right-pointers-in-each-node-ii