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
    
    /**
     * 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) {
            if(root == NULL)
              return ;
            TreeLinkNode *pleft=root->left,*pacross=pleft,*parent=root;
            root->next = NULL;
            while(pleft)
            {
                while(parent)
                {
                    if(pacross != pleft)
                    {
                        pacross->next = parent->left;
                        pacross = pacross->next;
                    
                    }
                    pacross->next = parent->right;
                    pacross = pacross->next;
                    parent = parent->next;
                }
                pacross->next = NULL;
                parent = pleft;
                pleft = pleft->left;
                pacross = pleft;
            }
              
        }
    };
  • 相关阅读:
    关于C语言中%p和%X的思考
    multimap员工分组案例
    set容器查找操作使用
    绘制漂亮的思维导图
    [deque容器练习]打分案例
    【LeetCode】1162. 地图分析
    【LeetCode】820. 单词的压缩编码
    【LeetCode】914. 卡牌分组
    【LeetCode】999. 车的可用捕获量
    【LeetCode】3. 无重复字符的最长子串
  • 原文地址:https://www.cnblogs.com/Xylophone/p/3812778.html
Copyright © 2011-2022 走看看