LeetCode: 116. Populating Next Right Pointers in Each Node

本文介绍如何解决LeetCode上的116题“填充每个节点的下一个右侧节点指针”。采用递归方法，通过分治思想实现完全二叉树中节点间的next指针连接。确保空间复杂度为O(1)。

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LeetCode: 116. Populating Next Right Pointers in Each Node

题目描述

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,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

题目大意：将给定的完全二叉树(所有的叶子节点在同一层，所有的父节点都有两个孩子)，连接成上图的形式(要求空间复杂度为 O(1))。

解题思路 —— 递归求解, 分治思想

给定的二叉树:
对左子树执行 connect 操作：
对右子树执行 connect 操作:
合并: 分别将左子树的每一层最右边的节点链接到右子树每一层的最左边节点:
链接第二层:

链接第三层:

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 == nullptr) return;

        // 分别对左右子树操作
        connect(root->left);
        connect(root->right);

        // 将左子树的每一层的最右边链接到右子树的最左边
        TreeLinkNode* leftTreeRight = root->left;
        TreeLinkNode* rightTreeLeft = root->right;
        while(leftTreeRight != nullptr && rightTreeLeft != nullptr)
        {
            leftTreeRight->next = rightTreeLeft;
            leftTreeRight = leftTreeRight->right;
            rightTreeLeft = rightTreeLeft->left;
        }
    }
};