leetcode-116. Populating Next Right Pointers in Each Node

leetcode-116. 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

这题是非常简单的题目。但是我给想复杂了。我其实是按照第117题做的。后来看到有个非常漂亮的答案。这里一个可以利用的点是踏实全二叉树。所以有固定的结构关系，没必要用队列或者栈去储存信息。所有就有了下面非常漂亮的解法。

答案来源

/**
* 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 pre = root;
TreeLinkNode cur = null;
while(pre.left!=null){
cur = pre;
while(cur!=null){
cur.left.next = cur.right;
if(cur.next!=null) cur.right.next = cur.next.left;
cur = cur.next;
}
pre = pre.left;
}
}
}