Wednesday, July 8, 2015

LeetCode [230] Kth Smallest Element in a BST

 230. Kth Smallest Element in a BST

Medium

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.
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
/**
 * 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:
    int kthSmallest(TreeNode* root, int k) {
        int nLeft = countN(root->left);
        if(k==nLeft+1) return root->val;
        else if(k<=nLeft) return kthSmallest(root->left, k);
        else return kthSmallest(root->right, k-nLeft-1);
    }
    
    int countN(TreeNode* root){
        if(!root) return 0;
        return 1+countN(root->left)+countN(root->right);
    }
};


 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
/**
 * 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:
    int kthSmallest(TreeNode* root, int k) {
        int ret, cur = k;
        dfs(root, cur, ret);
        return ret;
    }
    void dfs(TreeNode* root, int &cur, int &ret){
        if(!root) return;
        dfs(root->left, cur, ret);
        cur--;
        if(cur==0) ret = root->val;
        dfs(root->right, cur, ret);
    }
};


 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int left, ret;
    public int kthSmallest(TreeNode root, int k) {
        left = k;
        dfs(root);
        return ret;
    }
    
    void dfs(TreeNode node){
        if(node==null) return;
        dfs(node.left);
        left--;
        if(left==0) ret = node.val;
        dfs(node.right);
    }
}
Posted by at
Labels: ,

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)