首页 > 代码库 > Leetcode 树 Populating Next Right Pointers in Each Node
Leetcode 树 Populating Next Right Pointers in Each Node
本文为senlie原创,转载请保留此地址:http://blog.csdn.net/zhengsenlie
Populating Next Right Pointers in Each Node
Total Accepted: 13657 Total Submissions: 39540Given 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
题意:给定一棵perfect binary tree,将它每一个节点的next指针都指向该节点右边的节点
思路:dfs
在connect一棵树的时候,需要知道这棵树的根节点和它右边的节点
1.将树的根节点和它右边的节点连接起来
2.递归地将左子树connect起来,需要知道左子树节点和右子树节点
3.递归地将右子树connect起来,需要知道右子树节点和根右边的节点的左子树节点
递归函数为:
void connect(TreeLinkNode *root, TreeLinkNode *sibling)
表示将以root为根的树和它的兄弟树sibling连接起来
复杂度:时间O(n), 空间O(log n)
相关题目:
Populating Next Right Pointers in Each Node II
/** * 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){ connect(root, NULL); } void connect(TreeLinkNode *root, TreeLinkNode *sibling){ if(!root) return ; root->next = sibling; connect(root->left, root->right); connect(root->right, sibling ? sibling->left : NULL); } };