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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:
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
思路：
递归，对root的左右进行连接操作，然后先递归右树，再递归左树。
package tree; public class PopulatingNextRightPointersInEachNodeII { public void connect(TreeLinkNode root) { if (root == null || (root.left == null && root.right == null)) return; TreeLinkNode next = root.next; while (next != null) { if(next.left != null) { next = next.left; break; } if (next.right != null) { next = next.right; break; } next = next.next; } if (root.right != null) root.right.next = next; if (root.left != null) root.left.next = root.right == null ? next : root.right; connect(root.right); connect(root.left); } }
LeetCode - Populating Next Right Pointers in Each Node II
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原文地址：http://www.cnblogs.com/shuaiwhu/p/5127482.html
