230. Kth Smallest Element in a BST -M

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

 

Example 1:

Input: root = [3,1,4,null,2], k = 1
   3
  / \
 1   4
  \
   2
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
       5
      / \
     3   6
    / \
   2   4
  /
 1
Output: 3

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?

 

Constraints:

• The number of elements of the BST is between 1 to 10^4.
• You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

A:
就是每次计算左子树的大小---------------会有重复计算的问题--------------

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    int getCount(TreeNode* root){
        if(! root){
            return 0;
        }
        return 1 + getCount(root->left) + getCount(root->right);
    }
public:
    int kthSmallest(TreeNode* root, int k) {
        // we can assume root is not NULL
        int lCount = getCount(root->left);
        if(k <= lCount){
            return kthSmallest(root->left, k);
        }else if(k == lCount +1 ){
            return root->val;
        }else{
            return kthSmallest(root->right, k-lCount-1);
        }

    }
};

----------------------------------最多走一遍---------利用了 C++ 的参数 引用------------------

class Solution {

public:

int kthSmallest(TreeNode* root, int k) {

int res = -1;

helper(root, k, res);

return res;

}

private:

void helper(TreeNode* root, int& k, int& res) { // root is not NULL

if (res >= 0 || !root || k <= 0)

return;

helper(root->left, k, res);

k--; // count the node : root

if (k == 0) {

res = root->val;

return;

}

if (k >= 1)

helper(root->right, k, res);

}

};

************. 第三遍 ********改进：用返回值，代替了一个引用参数  ******

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        return helper(root, k);
    }
private:
    int helper(TreeNode* root, int& k) { // safely assume root is not NULL
        if (!root || k < 0)
            return INT_MIN;
        int l = helper(root->left, k);
        if(l >= 0)
            return l;
        k--; // count the node : root
        if (k == 0) {
            return root->val;
        }
        return helper(root->right, k);
    }
};

Mistakes:
1：  如果想通过监控res 是否是nullptr， 那么需要 double Pointer