zoukankan      html  css  js  c++  java
  • 【LeetCode】Populating Next Right Pointers in Each Node

    http://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/

    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

    /**
     * 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) {
            Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
            Queue<TreeLinkNode> tempqueue = new LinkedList<TreeLinkNode>();
            if(null==root)
                return;
            queue.offer(root);
            TreeLinkNode cur=root;
            while(!queue.isEmpty()||cur!=null){
                cur = queue.peek();
                queue.poll();
                if(cur.left!=null)
                    tempqueue.offer(cur.left);
                if(cur.right!=null)
                    tempqueue.offer(cur.right);
                
    
                if(queue.isEmpty()){
                    cur.next=null;
                    cur=null;
                    queue=tempqueue;
                    tempqueue=new LinkedList<TreeLinkNode>();
                }else{
                    cur.next=queue.peek();
                    cur=null;
                }
            }
        }
    }
  • 相关阅读:
    【算法•日更•第三十四期】最大流算法
    【算法•日更•第三十三期】网络流基础知识(最大流)
    【原】Java学习笔记017
    【原】Java学习笔记016
    【原】Java学习笔记015
    【原】Java学习笔记014
    【原】Java学习笔记013
    【原】Java学习笔记012
    【原】Java学习笔记011
    【原】Java学习笔记010
  • 原文地址:https://www.cnblogs.com/yixianyixian/p/3714658.html
Copyright © 2011-2022 走看看