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

    层次遍历

     1 /**
     2  * Definition for binary tree with next pointer.
     3  * struct TreeLinkNode {
     4  *  int val;
     5  *  TreeLinkNode *left, *right, *next;
     6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
     7  * };
     8  */
     9 class Solution {
    10 public:
    11     void connect(TreeLinkNode *root) {
    12        if (root == NULL) {
    13             return;
    14         }
    15         queue<TreeLinkNode* > q;
    16         TreeLinkNode *p;
    17         int idx = 1, n;
    18         q.push(root);
    19         while (!q.empty()) {
    20             n = idx - 1;
    21             idx = 0;
    22             p = q.front();
    23             if (q.front()->left != NULL) {
    24                 q.push(q.front()->left);
    25                 idx++;
    26             }
    27             if (q.front()->right != NULL) {
    28                 q.push(q.front()->right);
    29                 idx++;
    30             }
    31             q.pop();
    32             for (int i = 0; i < n; ++i) {
    33                 p->next = q.front();
    34                 if (q.front()->left != NULL) {
    35                     q.push(q.front()->left);
    36                     idx++;
    37                 }
    38                 if (q.front()->right != NULL) {
    39                     q.push(q.front()->right);
    40                     idx++;
    41                 }
    42                 p = p->next;
    43                 q.pop();
    44             }
    45             p->next = NULL;
    46         } 
    47     }
    48 };
  • 相关阅读:
    Redis到底该如何利用?
    AutoMapper搬运工之自定义类型转换
    AutoMapper搬运工之初探AutoMapper
    【ELK】docker-compose搭建ELK单机环境
    [flowable工作流引擎]基本概念及术语
    shell遍历文件夹读取文件夹下的文件
    vector类的简单实现
    string类的实现
    接口安全认证
    C# lock private VS private static
  • 原文地址:https://www.cnblogs.com/easonliu/p/3637003.html
Copyright © 2011-2022 走看看