LeetCode #116 - Populating Next Right Pointers in Each Node - Medium
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Problem
Given a binary treestruct 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).
Example
Given the following perfect binary tree, 1 / \ 2 3 / \ / \4 5 6 7After calling your function, the tree should look like: 1 -> NULL / \ 2 -> 3 -> NULL / \ / \4->5->6->7 -> NULL
Algorithm
整理一下题意:给定一棵二叉树,要求给每个节点添加一个next指针。next指针指向该节点的右边节点,若右边无节点则指向NULL。要求使用常数空间。
由于next指针指向右边的特性,可以确定使用BFS思路。但是题目要求常数空间,所以常规的递归DFS和队列BFS均不可使用,要如何实现呢?
如果假设某一层的next指针已经实现,那么可以直接利用next指针轻松实现层次内的转移。由于根节点的next指针必然指向NULL,那么这个思路是可行的。确定思路:利用next指针,且已知上一层可以推知下一层。
设置两层循环,第一层循环由本层起始节点转到下一层起始节点,第二层循环在本层内操作。注意到左子的next是右子,右子的next是父节点next的左子。若父节点无next,则右子next为NULL。
代码如下。
//BFSclass Solution {public: void connect(TreeLinkNode *root) { if(root==NULL) return; TreeLinkNode*node; while(root->left!=NULL){ node=root; while(node!=NULL){ node->left->next=node->right; if(node->next!=NULL) node->right->next=node->next->left; else node->right->next=NULL; node=node->next; } root=root->left; } }};
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