[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