Leetcode: Populating Next Right Pointer in Each Node
2013-07-12 11:22
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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:
要求常数存储空间,所以不能用层遍历
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
要求常数存储空间,所以不能用层遍历
/** * 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) { // Start typing your Java solution below // DO NOT write main() function while(root != null){ TreeLinkNode pre = root; TreeLinkNode sib; while(pre != null){ if(pre.left != null) pre.left.next = pre.right; sib = pre.next; if(sib != null && pre.right != null) pre.right.next = sib.left; pre = sib; } root = root.left; } } }
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