原题：
Given a binary tree
=>给定一个二叉树
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.
=>把每一个节点的next指针指向它右面的节点，假如右边没有节点，则指向null
Initially, all next pointers are set to NULL.
=>所有的next节点都是初始化为null的。
Note:
=>注意
- You may only use constant extra space.
- =>最好只使用有限的额外空间
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
- =>假设是一个完美的二叉树（就是所有的叶子都是在一个level上，所有的父子树都有两个子节点）
For example,
=>例如
Given the following perfect binary tree,
=>给定的完美二叉树
1
/ \
2 3
/ \ / \
4 5 6 7
After calling your function, the tree should look like:
=>在执行了函数之后，就如下图所示：
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ / \
4->5->6->7 -> NULL
/**
* 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) {
}
};晓东分析：
首先想到的就是二叉树的遍历，只要把先遍历到的左子树指向右子树，然后就可以组成如下的三角形,其中null是初始化的值。(5也是指向null的，未画出)
1 -> null
/ \
2 -> 3 ->null
/ \ / \
4->5 6->7->null
下面一步，就是把5和6之间联系起来，这就是右子树的next指向自己next的左子树。这样就可以得到
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ / \
4->5->6->7 -> NULL
下面就是各种遍历的算法选择了。选择了一个先序遍历，见http://blog.csdn.net/u011960402/article/details/14517135里面有详细描述。于是有了下面的代码
代码实现：
实现一：递归
/**
* 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 (NULL == root) {
return;
}
TreeLinkNode *left = root->left;
TreeLinkNode *right = root->right;
/* set the next pointer for his left child */
if (NULL != left) {
left->next = right;
}
/* set the next pointer for his right child */
if (NULL != right && NULL != root->next) {
right->next = root->next->left;
}
/* do it recursively */
connect(left);
connect(right);
}
};运行结果：
|
14 / 14 test cases passed.
|
Status: Accepted |
|
Runtime:
52 ms
|
实现二：
非递归实现：
/**
* 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;
stack<TreeLinkNode*> TreeStack;
while(root || !TreeStack.empty()){
while(root){
TreeStack.push(root);
if(root->left){
root->left->next = root->right;
if(root->next)
root->right->next = root->next->left;
}
root = root->left;
}
root = TreeStack.top();
TreeStack.pop();
root = root->right;
}
}
};运行结果：
|
14 / 14 test cases passed.
|
Status: Accepted |
|
Runtime:
104 ms
|
令人奇怪的是非递归反而比递归的算法慢了，比较奇怪。
若您有更好的算法可以提出，谢谢