LeetCode OJ - Populating Next Right Pointers in Each Node II

题目：

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

解题思路：

　　方法一：直接进行广度优先遍历，在遍历的过程中对next指针赋值。

　　方法二：可以利用生成的next指针来横向扫描，即得到一层的next指针之后，可以利用这一层的next指针来给下一层的next指针赋值。

代码：

　　方法一代码：

　　

/*** 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;}queue<TreeLinkNode*> one;queue<TreeLinkNode*> another;one.push(root);TreeLinkNode* cur;TreeLinkNode* next;while(!(one.empty() && another.empty())) {if (!one.empty()) {cur = one.front();one.pop();if (cur->left != NULL) another.push(cur->left);if (cur->right != NULL) another.push(cur->right);while (!one.empty()) {next = one.front();one.pop();if (next->left != NULL) another.push(next->left);if (next->right != NULL) another.push(next->right);cur->next = next;cur = next;} cur->next = NULL;}if (!another.empty()) {cur = another.front();another.pop();if (cur->left != NULL) one.push(cur->left);if (cur->right != NULL) one.push(cur->right);while (!another.empty()) {next = another.front();another.pop();if (next->left != NULL) one.push(next->left);if (next->right != NULL) one.push(next->right);cur->next = next;cur = next;} cur->next = NULL;}}}
};

方法二代码：

/*** 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:TreeLinkNode *findNext(TreeLinkNode *head){while(head != NULL && head->left == NULL && head->right == NULL)head = head->next;return head;}void connect(TreeLinkNode *root) {if(root == NULL) return;TreeLinkNode *head, *last, *nexhead;for(head = root; head != NULL; head = nexhead){head = findNext(head);if(head == NULL) break;if(head->left != NULL) nexhead = head->left;else nexhead = head->right;for(last = NULL; head != NULL; last = head, head = findNext(head->next)){if(head->left != NULL && head->right != NULL)head->left->next = head->right;if(last == NULL) continue;if(last->right != NULL) last->right->next = head->left != NULL ? head->left : head->right;else last->left->next = head->left != NULL ? head->left : head->right;}}}
};