[leedcode 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

/**
* 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) {
//注意：
//1、默认next为空，因此不用讨论最右侧节点的next问题（包括根节点）
//2、注意例子中5->6，求非最外侧的右节点的next，此时需要结合根节点是否有next进行判断
if(root==null) return;

if(root.left!=null){
root.left.next=root.right;
}
if(root.right!=null){
if(root.next!=null){
root.right.next=root.next.left;
}
}
connect(root.left);
connect(root.right);
}
}