首页 > 代码库 > Leetcode:Populating Next Right Pointers in Each Node II
Leetcode:Populating Next Right Pointers in Each Node II
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
分析：此题是Populating Next Right Pointers in Each Node的升级版，去掉了是perfect binary tree的条件，去掉这个条件带来的影响是我们必须查找节点p孩子的next和下一层的head节点，迭代代码如下：
class Solution {public: void connect(TreeLinkNode *root) { if(root == NULL) return; TreeLinkNode *head = root; while(head){ TreeLinkNode *cur = head; while(cur){ TreeLinkNode *next = NULL, *p; //find next node for(p = cur->next; p && p->left == NULL && p->right == NULL; p = p->next); if(p == NULL) next = NULL; else next = p->left?p->left:p->right; //level link if(cur->left){ cur->left->next = cur->right?cur->right:next; } if(cur->right) cur->right->next = next; cur = cur->next; } //find next head TreeLinkNode *q; for(q = head; q && q->left == NULL && q->right == NULL; q = q->next); if(q == NULL) head = NULL; else head = q->left?q->left:q->right; } }};
上述代码太繁琐，在上面代码的基础上，我们进行简化，实际上在每一层，我们只需维护两个变量，一个next指向下一层的第一个node，一个prev表示同一层的前一个节点。代码如下：
class Solution {public: void connect(TreeLinkNode *root) { while (root) { TreeLinkNode * next = nullptr; // the first node of next level TreeLinkNode * prev = nullptr; // previous node on the same level for (; root; root = root->next) { if (!next) next = root->left ? root->left : root->right; if (root->left) { if (prev) prev->next = root->left; prev = root->left; } if (root->right) { if (prev) prev->next = root->right; prev = root->right; } } root = next; // turn to next level } }};
Leetcode:Populating Next Right Pointers in Each Node II
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。