[LeetCode][Java] Populating Next Right Pointers in Each Node

题目：

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

题意：

给定一棵二叉树，最初每个节点的右指针都为空。现在要求将每个节点和其右指针相连，如果右指针不存在，指为NULL.最终的效果如上图所示.

算法分析：

一个递归就搞定了，就是递归让每一个节点他的左右子树通过next链接，直至到最后一层，然后递归左右节点，继续让他们的左右子树通过next链接。

代码如下：

public class Solution
{
public void connect(TreeLinkNode root)
{
if (root==null) return;
TreeLinkNode lr = root.left;
TreeLinkNode rl = root.right;

while(lr!=null && rl!=null)
{
lr.next = rl;
lr = lr.right;
rl = rl.left;
}
connect(root.left);
connect(root.right);
}
}