【LeetCode】Populating Next Right Pointers in Each Node Populating Next Right Pointers in Each Node II

参考链接

http://blog.csdn.net/beiyetengqing/article/details/8454924

题目描述

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 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

题目分析

给了一个完全二叉树，把所有在同一层的结点的next指针指向右侧结点

思路一：使用队列进行广度例行遍历。需要注意一个技巧就是如果记录一层结束，如代码里的levelSum的变化

思路二：使用递归。需要注意的技巧就是灵活使用已经求行的next指针，从左子树指向右子树。这种方法只适用于完整二叉树。

代码示例

#include
#include
using namespace std;

#define DEBUG

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

// 队列 广度遍历
#if 0
class Solution {
public:
void connect(TreeLinkNode *root) {
if(root == NULL)	return;
queue treeque;
treeque.push(root);
int levelSum = treeque.size();
while(!treeque.empty())
{
TreeLinkNode *front = treeque.front();
treeque.pop();
if(front->left != NULL)		treeque.push(front->left);
if(front->right != NULL)	treeque.push(front->right);

levelSum--;
if(levelSum == 0)
{
front->next=NULL;
levelSum = treeque.size();
#ifdef DEBUG
printf("%d next=NULL\n",front->val);
#endif
}
else
{
front->next = treeque.front();
#ifdef DEBUG
printf("%d next=%d\n",front->val,treeque.front()->val);
#endif
}
}
}
};
#elif 1
class Solution {
public:
void connect(TreeLinkNode *root) {
if(root == NULL)	return;
if(root->left != NULL)
root->left->next = root->right/*,printf("%d next=%d\n",root->left->val,root->right->val)*/;
if(root->right != NULL && root->next != NULL)//充分利用已经有的next指针
root->right->next = root->next->left/*,printf("%d next=%d\n",root->right->val,root->next->left->val)*/;

connect(root->left);
connect(root->right);
}
};
#endif
void test0()
{
TreeLinkNode node1(1);
TreeLinkNode node2(2);
TreeLinkNode node3(3);
TreeLinkNode node4(4);
TreeLinkNode node5(5);
TreeLinkNode node6(6);
TreeLinkNode node7(7);

node1.left = &node2;
node1.right = &node3;

node2.left = &node4;
node2.right = &node5;

node3.left = &node6;
node3.right = &node7;

Solution so;
so.connect(&node1);
}
int main()
{
test0();
return 0;
}

推荐学习C++的资料

C++标准函数库
http://download.csdn.net/detail/chinasnowwolf/7108919

在线C++API查询
http://www.cplusplus.com/