[leetcode 230]Kth Smallest Element in a BST

Given a binary search tree, write a function kthSmallest to find the kth 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?

Hint：

1、Try to utilize the property of a BST.
2、What if you could modify the BST node's structure?
3、The optimal runtime complexity is O(height of BST).

Credits:

Special thanks to @ts for adding this problem and creating all test cases.

给定一个顺序二叉树，找第K小的元素

第一次想到的就是中序遍历二叉树，得到第K小的元素，可是时间复杂度超过了O(height of BST).

可还是AC了

class Solution
{
public:

    void cal(TreeNode *root,vector<int> &temp)
    {
        if(root==NULL)
            return ;
        else
        {
            cal(root->left,temp);
            temp.push_back(root->val);
            cal(root->right,temp);
        }
    }

    int kthSmallest(TreeNode* root, int k)
    {
        vector<int> temp;
        cal(root,temp);
        return temp[k];
    }
};

顺便贴一个时间复杂度是 O(height of BST)的代码

class Solution {  
public:  
    int calcTreeSize(TreeNode* root){  
        if (root == NULL)  
            return 0;  
        return 1+calcTreeSize(root->left) + calcTreeSize(root->right);          
    }  
    int kthSmallest(TreeNode* root, int k) {  
        if (root == NULL)  
            return 0;  
        int leftSize = calcTreeSize(root->left);  
        if (k == leftSize+1){  
            return root->val;  
        }else if (leftSize >= k){  
            return kthSmallest(root->left,k);  
        }else{  
            return kthSmallest(root->right, k-leftSize-1);  
        }  
    }  
};  

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。

其他Leetcode题目AC代码：https://github.com/PoughER/leetcode