【leetcode】Populating Next Right Pointers in Each Node
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一、问题描述
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二、问题分析
首先明确是对二叉树的操作,而二叉树最常见的操作无非就是遍历,说到遍历无非就是前中后序遍历以及层次遍历。对应到题目中,我们首先考虑到的可能是层次遍历,然后把每一层直接链起来就行了,但层次遍历需要使用一个辅助的队列,而题目说You may only use constant extra space.显然有更好的方式。显然preOrderTraverse即可,但是遍历的时候我们需要注意,如果两个节点有相同的父节点那么很容易操作,反之呢,即两个节点没有相同的父节点但又形式上相邻,比如途中的5和6,仔细看图我们发现5和6的父节点是兄弟节点,即有next指针不为空。
三、Java AC代码
public void connect(TreeLinkNode root) { if (root==null) {return ;}if (root.left!=null && root.right!=null) {root.left.next=root.right;}if(root.next!=null && root.right!=null){ root.right.next = root.next.left;}connect(root.left);connect(root.right); }
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