LeetCode 116 Populating Next Right Pointers in Each Node
Given 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.
- Recursive approach is fine, implicit stack space does not count as extra space for this problem.
- 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 7
After calling your function, the tree should look like:
1 -> NULL / \ 2 -> 3 -> NULL / \ / \ 4->5->6->7 -> NULL
讲真这道题目废了我挺久时间的,是自己没考虑全面吧,写一波直接提交总是wa
c++
class Solution { public: void connect(TreeLinkNode *root) { if(root==NULL) return; queue<TreeLinkNode*> q; TreeLinkNode* pre = NULL; q.push(root); int i=0;int lever=0; int y = 0; while(!q.empty()) { TreeLinkNode* temp = q.front(); q.pop(); if(i==0||((i-y)==pow(2.0,lever))) { if(pre!=NULL) pre->next = temp; temp->next = NULL; y = i; lever++; } else{ if(i==y+1) {pre = temp; pre->next =NULL;} else { pre->next = temp;pre = temp;pre->next=NULL; } } i++; if(temp->left!=NULL) q.push(temp->left); if(temp->right!=NULL) q.push(temp->right); } } };