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LeetCode 108. Convert Sorted Array to Binary Search Tree （将有序数组转换成BST）
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108. Convert Sorted Array to Binary Search Tree
Given an array 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:
Given the sorted array: [-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
/**
* 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 {
public:
TreeNode* sortedArrayToBST(vector<int>& nums) {
if (nums.size() == ) return NULL;
if (nums.size() == )
{
return new TreeNode(nums[]);
} int mid = nums.size() / ;
TreeNode *root = new TreeNode(nums[mid]); vector<int> leftNs(nums.begin(), nums.begin() + mid);
vector<int> rightNs(nums.begin() + mid + , nums.end()); root->left = sortedArrayToBST(leftNs);
root->right = sortedArrayToBST(rightNs); return root;
}
};
法二：
/**
* 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 {
public:
TreeNode* sortedArrayToBST(vector<int>& nums) {
if (nums.size() == )
{
return nullptr;
} if(nums.size() == )
{
return new TreeNode(nums[]);
} int mid = ( + nums.size()) / ; TreeNode *root = new TreeNode(nums[mid]);
root->left = sortedArrayToBST(nums, , mid - );
root->right = sortedArrayToBST(nums, mid + , nums.size() - ); return root;
} TreeNode* sortedArrayToBST(vector<int>& nums, int start, int end){ if (start > end) return nullptr; int mid = (start + end) / ;
TreeNode *root = new TreeNode(nums[mid]);
root->left = sortedArrayToBST(nums, start, mid - );
root->right = sortedArrayToBST(nums, mid + , end); return root; } };
