LeetCode:Convert Sorted Array to Binary Search Tree,Convert Sorted List to Binary Search Tree

LeetCode:Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST

分析：找到数组的中间数据作为根节点，小于中间数据的数组来构造作为左子树，大于中间数据的数组来构造右子树，递归解法如下

/**
 * 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 *sortedArrayToBST(vector<int> &num) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int len = num.size();
        if(len == 0)return NULL;
        return sortedArrayToBSTRecur(num, 0, len-1);
    }
    TreeNode *sortedArrayToBSTRecur(vector<int> &num, int istart, int iend)
    {
        if(istart > iend)return NULL;
        int middle = (istart+iend)/2;
        TreeNode *res = new TreeNode(num[middle]);
        res->left = sortedArrayToBSTRecur(num, istart, middle-1);
        res->right = sortedArrayToBSTRecur(num, middle+1, iend);
        return res;
    }
};

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LeetCode: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

分析：和上一题同理，只不过要使用快慢指针来找到链表的中间节点                                                  本文地址

/**
 * 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 *sortedListToBST(ListNode *head) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(head == NULL)return NULL;
        ListNode *fast = head, *slow = head, *preSlow = NULL;
        while(fast->next && fast->next->next)
        {
            fast = fast->next->next;
            preSlow = slow;
            slow = slow->next;
        }
        TreeNode *res = new TreeNode(slow->val);
        fast = slow->next;
        delete slow;
        if(preSlow != NULL)
        {
            preSlow->next = NULL;
            res->left = sortedListToBST(head);
        }
        res->right = sortedListToBST(fast);
        return res;
    }
};

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