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社区首页 >专栏 >LeetCode 0109 - Convert Sorted List to Binary Search Tree

LeetCode 0109 - Convert Sorted List to Binary Search Tree

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Reck Zhang
发布2021-08-11 10:55:32
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发布2021-08-11 10:55:32
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文章被收录于专栏:Reck Zhang

Convert Sorted List to Binary Search Tree

Desicription

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

代码语言:javascript
复制
Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

Solution

代码语言:javascript
复制
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    TreeNode* buildTree(int left, int right, vector<int>& nums) {
        if(left > right)
            return 0;
        int mid = (left + right)>>1;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = buildTree(left, mid-1, nums);
        root->right = buildTree(mid+1, right, nums);
        return root;
    }
public:
    TreeNode* sortedListToBST(ListNode* head) {
    vector<int> nums;
        while(head) {
            nums.push_back(head->val);
            head = head->next;
        }
        return buildTree(0, nums.size()-1, nums);
    }
};
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  • Convert Sorted List to Binary Search Tree
    • Desicription
      • Solution
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