LeetCode 230: Kth Smallest Element in a BST

Given a binary search tree, write a function kthSmallest to find thekth smallest element in it.

Note: 
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

 

在二叉搜索树种，找到第K个元素。

算法如下：

1、计算左子树元素个数left。

2、 left+1 = K，则根节点即为第K个元素

      left >=k, 则第K个元素在左子树中，

     left +1 <k, 则转换为在右子树中，寻找第K-left-1元素。

代码如下：

/**
 * 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:
    //  计算左子树元素个数的函数
    int calculateSize(TreeNode* root){
        if(root == NULL)
            return 0;
        return 1 + calculateSize(root->left) + calculateSize(root->right);
    }
    int kthSmallest(TreeNode* root, int k) {
        if (root == NULL)
            return 0;
        int leftSize = calculateSize(root->left);   //计算左子树的元素个数
        if(k == leftSize+1) // 如果k == 左子树 + 1，那么正好的根节点，因为二叉搜索树的左子树小于根，右子树大于根
            return root->val;
        else if(k <= leftSize)  //left >=k, 则第K个元素在左子树中
            return kthSmallest(root->left, k);
        else                    //eft +1 <k, 则转换为在右子树中，寻找第K-left-1元素。
            return kthSmallest(root->right, k-leftSize-1);
        
    }
};