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
因此，对一个节点需要向右找到第一个节点。
对于left，如果right不存在，就在father的next节点去找left/right，依次找下去。
对于right，直接在father的next节点开始找。
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 TreeLinkNode node; 11 public void connect(TreeLinkNode root) { 12 if (root==null) { 13 return; 14 } 15 16 TreeLinkNode left = root.left; 17 TreeLinkNode right = root.right; 18 19 if (left != null) { 20 left.next = right; 21 if (right == null) { 22 node = root.next; 23 while (node != null && left.next == null) { 24 left.next = node.left; 25 if (left.next == null) { 26 left.next = node.right; 27 } 28 node = node.next; 29 } 30 } 31 } 32 33 if (right != null) { 34 node = root.next; 35 while (node != null && right.next == null) { 36 right.next = node.left; 37 if (right.next == null) { 38 right.next = node.right; 39 } 40 node = node.next; 41 } 42 } 43 44 connect(right); 45 connect(left); 46 } 47 }
LeetCode Populating Next Right Pointers in Each Node II
原文：http://www.cnblogs.com/birdhack/p/4158823.html