【LeetCode】117.Populating Next Right Pointers in Each Node II(Medium)解题报告
2018-04-03 21:09
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【LeetCode】117.Populating Next Right Pointers in Each Node II(Medium)解题报告
题目地址:https://leetcode.com/problems/unique-binary-search-trees/description/
题目描述:
Follow up for problem “Populating Next Right Pointers in Each Node”.
What if the given tree could be any binary tree? Would your previous solution still work?
Note:You may only use constant extra space.
Solution:
Date:2018年4月3日
题目地址:https://leetcode.com/problems/unique-binary-search-trees/description/
题目描述:
Follow up for problem “Populating Next Right Pointers in Each Node”.
What if the given tree could be any binary tree? Would your previous solution still work?
Note:You may only use constant extra space.
For example, Given the following binary tree, 1 / \ 2 3 / \ \ 4 5 7 After calling your function, the tree should look like: 1 -> NULL / \ 2 -> 3 -> NULL / \ \ 4-> 5 -> 7 -> NULL
Solution:
/** * Definition for binary tree with next pointer. * public class TreeLinkNode { * int val; * TreeLinkNode left, right, next; * TreeLinkNode(int x) { val = x; } * } 很重要 time : O(n) space : O(1) */ public class Solution { public void connect(TreeLinkNode root) { if(root == null) return; TreeLinkNode head = null; TreeLinkNode cur = root; TreeLinkNode pre = null; while(cur!=null){ while(cur!=null){ if(cur.left!=null){ if(pre!=null){ pre.next = cur.left; }else head = cur.left; pre = cur.left; } if(cur.right!=null){ if(pre!=null){ pre.next = cur.right; }else head = cur.right; pre = cur.right; } cur = cur.next; } cur = head; pre = null; head = null; } } }
Date:2018年4月3日
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