Populating Next Right Pointers in Each Node - LeetCode 116

题目描述：

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

Tag：Tree Depth-first Search

分析：方法1. 层次遍历。从根节点开始，其next必然为NULL，然后放入一个队列q中，接着循环处理q：

a.若q非空：

b. 若q非空，一次弹出每个元素m，将其的左右孩子无论空与否，依次放入一个向量vec中，直到q为空。

c. 遍历vec，将vec中的非空元素一次放入q中，同时修改它的next指针为它在vec中的下一个元素。

d. 继续过程a

以下是C++实现代码：

/**非递归//////32ms///*/
* 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;
TreeLinkNode *cur = root;
queue<TreeLinkNode* > q;
root->next = NULL; //修改root的next指针
if(cur != NULL)
{
q.push(cur);
while(!q.empty()) //此处判断是否最后一层处理完
{
int len = 0;
vector<TreeLinkNode*> vec; //存放即将要处理得层的每个节点的左右节点
while(!q.empty()) //此处判断某一层是否处理完
{
TreeLinkNode *cur = q.front();
q.pop();
vec.push_back(cur->left);  //依次将左右孩子放入vec中，无论空与否
len++;
vec.push_back(cur->right);
len++;
}
for(int i = 0; i < len-1; i++) //修改非空节点的next指针
{
if(vec[i] != NULL)
{
q.push(vec[i]);
vec[i]->next = vec[i+1];
}
}
if(vec[len-1] != NULL) //处理每层的最后一个元素的next指针
{
q.push(vec[len-1]);
vec[len - 1]->next = NULL;
}
}
}
}
};