116.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 nonext right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

·        You may only useconstant extra space.

·        You may assumethat it is a perfect binary tree (ie, all leaves are at the same level, andevery 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

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#pragma once
#include<iostream>
#include<queue>
using namespace std;

struct TreeLinkNode {
	int val;
	TreeLinkNode *left, *right, *next;
	TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
};

void connect(TreeLinkNode *root) {
	if (!root)
		return;
	TreeLinkNode* p;
	queue<TreeLinkNode*> queue;
	queue.push(root);
	int level = 0;//记录当前第几层
	int count = 0;//记录当前是该层的第几个节点，通过与pow（2,level）的大小关系来决定是否设置next指针
	while (!queue.empty())
	{
		count++;
		p = queue.front();
		queue.pop();
		if (p->left)//满二叉树，只需判断一个孩子即可
		{
			queue.push(p->left);
			queue.push(p->right);
		}
		if (count < (int)pow(2, level))//同一层中，设置next指针
			p->next = queue.front();
		else//下一层
		{
			count = 0;
			level++;
		}
	}
}

void main()
{
	system("pause");
}