【LeetCode】108. Convert Sorted Array to Binary Search Tree

Problem:

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

Solution:

复习一下BST的基本性质：简而言之，对于BST中任意结点x，若其有左右结点 l 或 r ，需满足 l.key ≤ x.key ≤ r.key

BST的性能（基本操作耗时）与树高成正比，可改进为较为高效的平衡二叉树，即保持整棵树左右均匀，任意左右子树高度差不大于1

如题将有序数组转化为二叉平衡树，可将数组从中间分成左右两部分，在分别对中间根节点赋值，左子树和右子树的确定用递归方法

Code:

//in C language

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
 struct TreeNode *createTree(int left,int right, int nums[]){
    if(right-left<0)
        return NULL;
    //if负责终止递归
    int mid=(left+right)/2;
    struct TreeNode *root=(struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val=nums[mid];
    root->left=createTree(left,mid-1,nums);
    root->right=createTree(mid+1,right,nums);
    
    return root;
}
 
struct TreeNode* sortedArrayToBST(int* nums, int numsSize) {
    return createTree(0, numsSize-1, nums);
}

ps:似乎是谷哥哥家的题目～加油！