leetcode:Populating Next Right Pointers in Each Node
2014-12-22 19:39
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题目描述:
Given a binary tree
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
Initially, all next pointers are set to
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,
After calling your function, the tree should look like:
不得不说:刚开始先是想到了用队列层次遍历,后来发现还是有更优先的代码;
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
不得不说:刚开始先是想到了用队列层次遍历,后来发现还是有更优先的代码;
/** * 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) { TreeLinkNode level_start=root; while(level_start!=null){ TreeLinkNode cur=level_start; while(cur!=null){ if(cur.left!=null) cur.left.next=cur.right; if(cur.right!=null && cur.next!=null) cur.right.next=cur.next.left; cur=cur.next; } level_start=level_start.left; } } }
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