Populating Next Right Pointers in Each Node II

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

 

For example,
Given the following binary tree,

         1

       /  \

      2    3

     / \    \

    4   5    7

 

After calling your function, the tree should look like:

         1 -> NULL

       /  \

      2 -> 3 -> NULL

     / \    \

    4-> 5 -> 7 -> NULL

C++实现代码（使用层次遍历的方法，每次记录一层，对每一层的next进行赋值）：

#include<iostream>

#include<new>

#include<vector>

using namespace std;





// 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)

            return;

        vector<TreeLinkNode*> q;

        vector<TreeLinkNode*> curq;

        q.push_back(root);

        TreeLinkNode *tmp;

        while(!q.empty())

        {

            curq=q;

            q.clear();

            for(size_t i=0;i<curq.size();i++)

            {

                tmp=curq[i];

                if(i<curq.size()-1)

                    tmp->next=curq[i+1];

                else

                    tmp->next=NULL;

                if(tmp->left)

                    q.push_back(tmp->left);

                if(tmp->right)

                    q.push_back(tmp->right);

            }

        }

    }

    void createTree(TreeLinkNode *&root)

    {

        int i;

        cin>>i;

        if(i!=0)

        {

            root=new TreeLinkNode(i);

            if(root==NULL)

                return;

            createTree(root->left);

            createTree(root->right);

        }

    }

};



int main()

{

    Solution s;

    TreeLinkNode *root;

    s.createTree(root);

    s.connect(root);

}