zoukankan      html  css  js  c++  java
  • [LeetCode] 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
    

     解题思路:

    按层遍历一遍所有结点即可,每一次从左到右枚举一次当前层的节点(这一层的节点已经连上),如果下一层为空,则退出,否则顺序地将下一层连起来。

    /**
     * Definition for binary tree with next pointer.
     * struct TreeLinkNode {
     *  int val;
     *  TreeLinkNode *left, *right, *next;
     *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
     * };
     */
    class Solution {
    public:
        void connect(TreeLinkNode *root) {
            // IMPORTANT: Please reset any member data you declared, as
            // the same Solution instance will be reused for each test case.
        if(root == NULL)
            return;
            
        TreeLinkNode *head = root, *last = root, *tmp_last = root;//store the horizontal list
        root -> next = NULL;
        
        //if head -> left == NULL, then this is the leaf level, end
        while(head -> left != NULL)
        {
            //list the head level node
            last = head -> left;//last stores the next level, which is what we want to connect
            last -> next = head -> right;//connect the right to the left
            last = head -> right;//last node switch to the right node of the head
            tmp_last = head;//head level last node
            while(tmp_last -> next != NULL)
            {
                tmp_last = tmp_last -> next;
                last -> next = tmp_last -> left;
                (tmp_last -> left) -> next = tmp_last -> right;
                last = tmp_last -> right;
            }
            
            last -> next = NULL;
            head = head -> left;
        } 
        }
    };
  • 相关阅读:
    WinMain函数的修饰符WINAPI的含义
    java字节码指令集简介
    vs2010里面的ipch文件和.sdf文件是什么
    java查看class字节码文件
    从汇编看c++的extern关键字
    highcharts系列教程
    highcharts的文档介绍(英文)
    关于firebug中行号和源文件不一致的问题
    ios中的流状态的定义
    highcharts翻译系列
  • 原文地址:https://www.cnblogs.com/changchengxiao/p/3415988.html
Copyright © 2011-2022 走看看