230. Kth Smallest Element in a BST

Problem Description

Given the root of a binary search tree, and an integer k, return the kth smallest value (1-indexed) of all the values of the nodes in the tree.

Example 1:

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

Example 2:

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

Constraints:

• The number of nodes in the tree is n.
• 1 <= k <= n <= 10⁴
• 0 <= Node.val <= 10⁴

Follow-up:

If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize the solution?

Python Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
        # Iterative inorder traversal
        stack = []
        curr = root
        
        while curr or stack:
            # Traverse to leftmost node
            while curr:
                stack.append(curr)
                curr = curr.left
            
            # Process current node
            curr = stack.pop()
            k -= 1
            if k == 0:
                return curr.val
            
            # Move to right subtree
            curr = curr.right

Java Solution

/**
 * 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 {
    public int kthSmallest(TreeNode root, int k) {
        // Iterative inorder traversal
        Stack stack = new Stack<>();
        TreeNode curr = root;
        
        while (curr != null || !stack.isEmpty()) {
            // Traverse to leftmost node
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            
            // Process current node
            curr = stack.pop();
            k--;
            if (k == 0) {
                return curr.val;
            }
            
            // Move to right subtree
            curr = curr.right;
        }
        
        return -1; // Should never reach here if k is valid
    }
}

C++ Solution

/**
 * 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) {
        // Iterative inorder traversal
        stack stack;
        TreeNode* curr = root;
        
        while (curr || !stack.empty()) {
            // Traverse to leftmost node
            while (curr) {
                stack.push(curr);
                curr = curr->left;
            }
            
            // Process current node
            curr = stack.top();
            stack.pop();
            k--;
            if (k == 0) {
                return curr->val;
            }
            
            // Move to right subtree
            curr = curr->right;
        }
        
        return -1; // Should never reach here if k is valid
    }
};

JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} k
 * @return {number}
 */
var kthSmallest = function(root, k) {
    // Iterative inorder traversal
    const stack = [];
    let curr = root;
    
    while (curr || stack.length) {
        // Traverse to leftmost node
        while (curr) {
            stack.push(curr);
            curr = curr.left;
        }
        
        // Process current node
        curr = stack.pop();
        k--;
        if (k === 0) {
            return curr.val;
        }
        
        // Move to right subtree
        curr = curr.right;
    }
    
    return -1; // Should never reach here if k is valid
};

C# Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int KthSmallest(TreeNode root, int k) {
        // Iterative inorder traversal
        var stack = new Stack();
        TreeNode curr = root;
        
        while (curr != null || stack.Count > 0) {
            // Traverse to leftmost node
            while (curr != null) {
                stack.Push(curr);
                curr = curr.left;
            }
            
            // Process current node
            curr = stack.Pop();
            k--;
            if (k == 0) {
                return curr.val;
            }
            
            // Move to right subtree
            curr = curr.right;
        }
        
        return -1; // Should never reach here if k is valid
    }
}