leetcode 117. Populating Next Right Pointers in Each Node II BFS广度优先遍历

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

he上一道题一样，代码一样使用，就是做一个简单的BFS广度优先遍历。

代码如下：

import java.util.ArrayList;import java.util.List;/*class TreeLinkNode {      int val;      TreeLinkNode left, right, next;      TreeLinkNode(int x) { val = x; }}*//* * 就是一个BFS广度优先遍历 * */public class Solution {    public void connect(TreeLinkNode root)     {        if(root == null)            return;        List<TreeLinkNode> myQueue = new ArrayList<TreeLinkNode>();        myQueue.add(root);        int count = 1;        while(myQueue.isEmpty()==false)        {            int tmp = 0;            for(int i=0;i<count-1;i++)            {                if(myQueue.get(0).left!=null)                {                    myQueue.add(myQueue.get(0).left);                    tmp++;                }                if(myQueue.get(0).right!=null)                {                    myQueue.add(myQueue.get(0).right);                    tmp++;                }                myQueue.get(0).next = myQueue.get(1);                myQueue.remove(0);            }                           if(myQueue.get(0).left!=null)            {                myQueue.add(myQueue.get(0).left);                tmp++;            }            if(myQueue.get(0).right!=null)            {                myQueue.add(myQueue.get(0).right);                tmp++;            }            myQueue.get(0).next = null;            myQueue.remove(0);            count = tmp;        }    }           }

下面是C++的做法，和上一道题一样，就是做一个BFS广度优先遍历

代码如下：

#include <iostream>#include <string>#include <vector>#include <queue>using namespace std;/*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)     {        queue<TreeLinkNode*> que;        if (root == NULL)            return;        que.push(root);        while (que.empty() == false)        {            int size = que.size();            vector<TreeLinkNode*> one;            for (int i = 0; i < size; i++)            {                TreeLinkNode* top = que.front();                que.pop();                one.push_back(top);                if (top->left != NULL)                    que.push(top->left);                if (top->right != NULL)                    que.push(top->right);            }            for (int i = 0; i <= (int)one.size() - 2; i++)                one[i]->next = one[i + 1];            one[one.size() - 1]->next = NULL;        }    }};