题目描述

题目链接：116. Populating Next Right Pointers in Each Node

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

struct Node { int val; Node *left; Node *right; Node *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.

例子

例子 1

Input: root = [1,2,3,4,5,6,7] Output: [1,#,2,3,#,4,5,6,7,#] Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with ‘#’ signifying the end of each level.

Follow Up

You may only use constant extra space. Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.

Note

Constraints

The number of nodes in the given tree is less than 4096. -1000 <= node.val <= 1000

解题思路

如果用层序遍历的话这道题的解法很直观，直接层序遍历输出每一层，并且让每层节点的 next 指向下一个即可，但是题目要求不使用额外空间。所以需要考虑其他方法，考虑到输入是一个完美二叉树，因此我们我们可以得知对于任何两个相邻的节点，他们的父节点要么是同一个，要么是也是相邻的，结合这个思路我们可以使用一个递归函数 connect(left, right) 的做法，思路如下：

• 对于根节点 root, 调用 connect(root->left, root->right)
• connect 函数中：
  • 首先连接输入的两个节点： left->next = right
  • 然后执行以下操作即可：
    • connect(left->left, left->right)
    • connect(left->right, right->left)
    • connect(right->left, right->right)

代码如下：

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root) {
        if (root == nullptr) return root;
        connect(root->left, root->right);
        return root;
        
    }
    
private:
    void connect(Node* left, Node* right) {
        if (left == nullptr || right == nullptr) return;
        left->next = right;
        connect(left->left, left->right);
        connect(left->right, right->left);
        connect(right->left, right->right);
    }
};

• 时间复杂度: O(n)
• 空间复杂度: O(h) – 递归深度

GitHub 代码同步地址： 116.PopulatingNextRightPointersInEachNode.cpp

其他题目： GitHub: Leetcode-C++-Solution 博客： Leetcode-Solutions