【leetcode】PopulatingNextRightPointersinEachNode
Question :
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
Anwser 1:
Travesal
/** * 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) { // Start typing your C/C++ solution below // DO NOT write int main() function if(NULL == root){ return; } TreeLinkNode *left = root; while(left && left->left && left->right) { root = left; while(root) { root->left->next = root->right; // connect two nodes of root if(root->next){ root->right->next = root->next->left; // connect two isolated nodes } root=root->next; // traversal the same level line } left=left->left; } } };
Anwser 2:
Recursive
/** * 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) { // Start typing your C/C++ solution below // DO NOT write int main() function if(NULL == root){ return; } if(root->left){ root->left->next = root->right; } if(root->right){ root->right->next = root->next ? root->next->left : NULL; } connect(root->left); connect(root->right); } };
Anwser 3:
Queue
/** * 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) { // Start typing your C/C++ solution below // DO NOT write int main() function if(NULL == root){ return; } queue<TreeLinkNode *> Q; if(root != NULL){ Q.push(root); } int row = 1; int count = 0; while(!Q.empty()){ TreeLinkNode *tmp = Q.front(); Q.pop(); count++; if(tmp->left) Q.push(tmp->left); if(tmp->right) Q.push(tmp->right); if(count == row){ tmp->next = NULL; count = 0; row *= 2; } else { tmp->next = Q.front(); } } } };
注意点:
1) Queue > Traversal > Recursive
2) Queue 没有递归的层次限制,可以使用很大的二叉树
版权所有: 本文系米扑博客原创、转载、摘录,或修订后发表,最后更新于 2013-04-13 22:19:43
侵权处理: 本个人博客,不盈利,若侵犯了您的作品权,请联系博主删除,莫恶意,索钱财,感谢!
转载注明: 【leetcode】PopulatingNextRightPointersinEachNode (米扑博客)