Populating Next Right Pointers in Each Node 题解

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题目来源：

Description

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).

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

Solution

class Solution {private:    void connectNode(vector
    
     & v) {        int size = v.size();        for (int i = 0; i <= size - 2; i++) {            v[i] -> next = v[i + 1];        }    }public:    void connect(TreeLinkNode *root) {        if (root == NULL)            return;        int levelNodeNum = 1;        int curLevelNodeNum = 0;        queue
     
       q;        vector
      
        v;        q.push(root);        while (!q.empty()) {            TreeLinkNode* node = q.front();            q.pop();            v.push_back(node);            if (node -> left != NULL)                q.push(node -> left);            if (node -> right != NULL)                q.push(node -> right);            curLevelNodeNum++;            if (curLevelNodeNum == levelNodeNum) {                levelNodeNum *= 2;                curLevelNodeNum = 0;                connectNode(v);                v.clear();            }        }    }};

解题描述

这道题是关于二叉树层次遍历问题的变种。题目给出的二叉树是完全二叉树，所以可以提前算出每一层的节点数目，因此来说还是相对比较容易的。所以基本的解决办法是，使用一个队列来存放节点。最开始将根节点加入队列。每次从队首取出一个节点，将其子节点加入队尾。然后使用一个计数变量来计算当前层次上已经加入队列的节点数目。一旦达到该层次的节点数目总和就对该层的节点进行next连接。

转载于:https://www.cnblogs.com/yanhewu/p/8041064.html

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