Populating Next Right Pointers in Each Node
Total Accepted: 72323 Total Submissions: 199207 Difficulty: Medium
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.
- 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
/** * 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 dfs(TreeLinkNode* root1,TreeLinkNode* root2){ if(!root1) return; root1->next = root2; dfs(root1->left,root1->right); if(!root2) return; dfs(root1->right,root2->left); dfs(root2->left,root2->right); } void connect(TreeLinkNode *root) { if(!root) return ; dfs(root->left,root->right); } };
Next challenges: (H) Populating Next Right Pointers in Each Node II
Populating Next Right Pointers in Each Node II
Total Accepted: 51329 Total Submissions: 158800 Difficulty: Hard
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
广度优先遍历
/** * 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) { if(!root) return ; TreeLinkNode* head = NULL; TreeLinkNode* tail = NULL; TreeLinkNode* cur = root; while(cur){ if(cur->left){ if(tail){ tail->next = cur->left; tail = cur->left; }else{ tail = cur->left; head = cur->left; } } if(cur->right){ if(tail){ tail->next = cur->right; tail = cur->right; }else{ tail = cur->right; head = cur->right; } } cur = cur->next; if(!cur){ cur = head; tail = NULL; head = NULL; } } } };