LeetCode OJ 116. 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

对于这个题目我们可以很方便的采用层序遍历来解决，每次遍历一层把下一层的节点依次连接起来。由于每一层相当于形成了一个链表，树的最左边的节点就是每一层的起始节点。我们遍历这个链表，把链表中每一个节点的左右节点相连，然后把链表中相邻节点的左孩子和右孩子相连即可。

代码如下：

/**
* 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) return;
TreeLinkNode temp = root;
if(temp.left!=null){
while(temp!=null){　　　　//遍历一层
temp.left.next = temp.right;
if(temp.next!=null){
temp.right.next = temp.next.left;
}
else{
temp.right.next = null;
}
temp = temp.next;
}
connect(root.left);　　　　//遍历下一层
}
return;
}
}