AppIntent论文阅读
Populating Next Right Pointers in Each Node
Total Accepted: 6129 Total Submissions: 17922My SubmissionsGiven 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
直接上递归法
void connect(TreeLinkNode *root) { if (!root || !root->left) return; root->left->next = root->right; con(root->left, root->right); } void con(TreeLinkNode *lTree, TreeLinkNode *rTree) { if (lTree->left) lTree->left->next = lTree->right; if (rTree->left) rTree->left->next = rTree->right; else return; lTree->right->next = rTree->left; con(lTree->left, lTree->right); con(lTree->right, rTree->left); con(rTree->left, rTree->right); }
下面参考leetcode论坛上的程序写了个程序
//新知识点: //重点注意:利用新构造的数据结构 void connect(TreeLinkNode* root) { if (!root || !root->left || !root->right) return; TreeLinkNode* rightSibling; TreeLinkNode* p1 = root; while (p1) { rightSibling = p1->next? p1->next->left:NULL; p1->left->next = p1->right; p1->right->next = rightSibling; p1 = p1->next; } connect(root->left); }
有人会觉得使用递归并不符合常数空间的题意,那么可以改成非递归:
void connect(TreeLinkNode *root) { TreeLinkNode *nextLev = root? root->left:NULL; while (nextLev) { for ( ; root; root = root->next) { TreeLinkNode *rightSibling = root->next? root->next->left:NULL; root->left->next = root->right; root->right->next = rightSibling; } root = nextLev; nextLev = nextLev->left; } }
郑重声明:本站内容如果来自互联网及其他传播媒体,其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享,并不代表本站赞同其观点和对其真实性负责,也不构成任何其他建议。