[LeetCode]Populating Next Right Pointers in Each Node

最新推荐文章于 2021-09-24 12:05:40 发布

风啸葛溪

最新推荐文章于 2021-09-24 12:05:40 发布

阅读量600

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode解题

本文链接：https://blog.youkuaiyun.com/Roger__King/article/details/23207315

LeetCode解题专栏收录该内容

28 篇文章

订阅专栏

本文介绍了一种填充二叉树中每个节点的next指针的方法，使其指向同一层级的下一个节点。提供了三种不同的C++实现方案，包括递归与迭代的方式，并详细解释了每种方法的工作原理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

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

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

思路：

C++ AC代码：

/**
 * 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 || root->left == NULL || root->right == NULL)
        return ;
        TreeLinkNode *p =root ;
        while  ( p!= NULL){
            p->left->next = p->right ;
            if ( p->next) 
            p->right->next =p->next->left;
            p = p->next;
        }
        connect(root->left);
    }
};

在网上看到另外两种解法：

//http://blog.youkuaiyun.com/njust_ecjtu/article/details/16361151

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 *left,*right;
       left = root->left;
       right = root->right;
       
       while(left&&right)//将俩子树间的空隙链接起来---》关键所在
       {
           left->next = right;
           
           left = left->right;//走向下一层
           right = right->left;
       }
       
       connect(root->left);//遍历左边
       connect(root->right);//遍历右边
        
    }

public:
     void connect(TreeLinkNode *root) {
                   connect(root, NULL);
                  }
private:
     void connect(TreeLinkNode *root, TreeLinkNode *sibling) {
                         if (root == nullptr)
                            return;
                        else
                            root->next = sibling;
  
                       connect(root->left, root->right);
                       if (sibling)
                          connect(root->right, sibling->left);
                     else
                           connect(root->right, nullptr);
               }
};