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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轉載于:https://www.cnblogs.com/dongguangqing/p/3727925.html

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