Populating Next Right Pointers in Each Node

Given a binary tree

struct TreeLinkNode {      TreeLinkNode *left;      TreeLinkNode *right;      TreeLinkNode *next;    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1       /  \      2    3     / \  / \    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL       /  \      2 -> 3 -> NULL     / \  / \    4->5->6->7 -> NULL

分析：递归

代码：

/** * Definition for binary tree with next pointer. * public class TreeLinkNode { *     int val; *     TreeLinkNode left, right, next; *     TreeLinkNode(int x) { val = x; } * } */public class Solution {     public void connect(TreeLinkNode root) {        if(root == null || root.left == null){            // 空树或者只有一个结点的树直接返回，由于是完美二叉树，因此只用判断左子树            return;        }                //分别先对左子树和右子树进行处理        connect(root.left);        connect(root.right);                //逐层连接左子树和右子树        TreeLinkNode leftNode = root.left;        TreeLinkNode rightNode = root.right;                while (leftNode != null){            leftNode.next = rightNode;            leftNode = leftNode.right;            rightNode = rightNode.left;        }            }}