[LeetCode 题解]:Convert Sorted List to Binary Search Tree
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
题意:给定一个有序的链表,将其转换成平衡二叉搜索树
思路: 二分法
要构建一个平衡二叉树,二分法无疑是合适的,至于如何分是的代码简洁,就需要用到递归了。
class Solution {
public:
// find middle element of the list
ListNode *getmiddleList(ListNode *left,ListNode *right){
//omit the condition : left!=right && left->next!=right
ListNode *pre,*last;
pre=left; last =left->next;
while(last!=right){
last = last->next;
if(last!=right){
last = last->next;
pre=pre->next;
}
}
return pre;
} // retri-BST constructor
TreeNode *getBST(ListNode *left,ListNode *right){
TreeNode *root = new TreeNode();
//no leaf
if(left==right) return NULL;
// only one leaf
if(left->next == right){
root->val=left->val;
return root;
}
//more than one leaf
ListNode *middle =getmiddleList(left,right);
root->val = middle->val;
root->left = getBST(left, middle);
root->right = getBST(middle->next,right);
return root;
}
TreeNode *sortedListToBST(ListNode *head) {
TreeNode* root= new TreeNode();
if(head==NULL) return NULL;
if(head->next==NULL){
root->val=head->val;
root->left=root->right=NULL;
return root;
}
ListNode *left,*middle,*right;
middle=left=head;
right=head->next;
while(right){
right=right->next;
if(right){
right=right->next;
middle=middle->next;
}
}
root->val=middle->val;
root->left = getBST(left, middle);
root->right= getBST(middle->next,right);
return root;
}
};
转载请注明出处:http://www.cnblogs.com/double-win/ 谢谢!
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