109 - Convert Sorted List to Binary Search Tree

最新推荐文章于 2021-11-02 12:49:42 发布

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本文详细介绍了如何通过计算链表长度并递归地找到中间节点来将排序单链表转换为高度平衡的二叉搜索树（BST）。此过程涉及链表操作、递归技巧和理解BST的性质。

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

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思路分析：

这题的关键是能找出当前链表的中间节点，然后再递归左右的子链表，开始的时候程序先计算链表总厂，然后传入两个前后索引指针，最后每次递归找出中间节点即可。

/**
 * 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:
    int calLen(ListNode *node)
    {
        int len = 0;
        while(node)
        {
            len++;
            node = node->next;
        }
        return len;
    }
    
    TreeNode *createTree(ListNode *node, int left, int right)
    {
        if (left > right)
            return NULL;
            
        int mid = (left + right) / 2;
        
        ListNode *p = node;
        
        for(int i = left; i < mid; i++)
            p = p->next;
            
        TreeNode *leftNode = createTree(node, left, mid - 1);
        TreeNode *rightNode = createTree(p->next, mid + 1, right);
        
        TreeNode *tNode = new TreeNode(p->val);
        
        tNode->left = leftNode;
        tNode->right = rightNode;
        
        return tNode;        
    }
    
    TreeNode *sortedListToBST(ListNode *head) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int len = calLen(head);
        return createTree(head, 0, len - 1);
    }
};