【easy】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.
//思路还是很清晰的:就是要取中间节点作为根节点,其他的分别是左子树和右子树,递归
/**
* 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) {
TreeNode *node = createBST(nums,,nums.size()-);
return node;
}
public:
TreeNode* createBST(vector<int>nums, int left, int right){
if (left>right)
return NULL;//??
int mid = (left+right)/;
TreeNode *node = new TreeNode(nums[mid]);
node->left = createBST(nums, left, mid - );
node->right = createBST(nums, mid+, right);
return node;
}
};
递归出口NULL注意,最终只是要返回根节点。
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