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:
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
Solution:
http://blog.csdn.net/wzy_1988/article/details/17412025
这道题目首要是找到右孩子的第一个有效的next链接节点，然后再处理左孩子。然后依次递归处理右孩子，左孩子
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 TreeLinkNode p = root.next; 14 while (p != null) { 15 if(p.left!=null){ 16 p=p.left; 17 break; 18 } 19 if(p.right!=null){ 20 p=p=p.right; 21 break; 22 } 23 p=p.next; 24 } 25 if (root.right != null) { 26 root.right.next = p; 27 } 28 if (root.left != null) { 29 root.left.next = (root.right == null) ? p : root.right; 30 } 31 connect(root.right); 32 connect(root.left); 33 } 34 }
[Leetcode] Populating Next Right Pointers in Each Node II
原文：http://www.cnblogs.com/Phoebe815/p/4099542.html