leetcode 116. Populating Next Right Pointers in Each Node BFS广度优先遍历

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,

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

题意很简单，就是做一个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，就是做一个BFS广度优先遍历即可

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;
}
}
};