LeetCode: Convert Sorted List to Binary Search Tree

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#tree #list #search #null #struct #function

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本文介绍了一种通过递归方法将有序链表转换为高度平衡二叉搜索树的技术，时间复杂度为O(n)，并详细解释了平衡二叉搜索树的概念及其构建过程。

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Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

For example,
Given list {9,12,14,17,19,23,50,54,67,72,76},
the below BST is one possible solution.

An example of a height-balanced tree. A height-balanced tree is a tree whose subtrees differ in height by no more than one and the subtrees are height-balanced, too.

利用递归，从低向上建立平衡二叉搜索树，时间复杂度为O(n)。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *convert(ListNode* &head, int nStart, int nEnd)
    {
        if (nStart > nEnd) return NULL;
        
        int mid = nStart + (nEnd-nStart)/2;
        TreeNode* left = convert(head, nStart, mid-1);
        TreeNode* parent = new TreeNode(head->val);
        parent->left = left;
        head = head->next;
        parent->right = convert(head, mid+1, nEnd);
        return parent;
    }
    
    TreeNode *sortedListToBST(ListNode *head) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL) return NULL;
        int nSize = 1;
        ListNode* ptr = head;
        while (ptr->next != NULL)
        {
            ptr = ptr->next;
            ++nSize;
        }
        return convert(head, 0, nSize-1);
    }
};