LeetCode OJ - Populating Next Right Pointers in Each Node II

題目：

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.

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For example,
Given the following binary tree,

         1/  \2    3/ \    \4   5    7

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After calling your function, the tree should look like:

         1 -> NULL/  \2 -> 3 -> NULL/ \    \4-> 5 -> 7 -> NULL

解題思路：

　　方法一：直接進行廣度優先遍歷，在遍歷的過程中對next指針賦值。

　　方法二：可以利用生成的next指針來橫向掃描，即得到一層的next指針之后，可以利用這一層的next指針來給下一層的next指針賦值。

代碼：

　　方法一代碼：

　　

/*** 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;}}}
};

方法二代碼：

/*** 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;}}}
};

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