LeetCode-Populating Next Right Pointers in Each Node II
2013-09-17 18:30
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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,
After calling your function, the tree should look like:
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
/** * 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: int Count(TreeLinkNode *root){ if(root==NULL)return 0; else return Count(root->left)+Count(root->right)+1; } void connect(TreeLinkNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function if(root==NULL)return; int count=Count(root); vector<TreeLinkNode*> queue; queue.resize(count); int head=0,tail=1; queue[0]=root; int current=1; int next=0; while(head<count){ if(queue[head]->left!=NULL){ queue[tail]=queue[head]->left; tail++; next++; } if(queue[head]->right!=NULL){ queue[tail]=queue[head]->right; tail++; next++; } current--; if(current==0){ current=next; next=0; } else{ queue[head]->next=queue[head+1]; } head++; } } };
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