230. Kth Smallest Element in a BST

MediumTreeDepth-First SearchBinary Search TreeBinary Tree

Problem Description

The problem asks for the kth smallest value in a Binary Search Tree (BST). A BST is a tree data structure where each node has at most two children, known as the left child and the right child. In a BST, the left child's value is less than its parent's value, and the right child's value is greater than its parent's value. This property of BST provides an in-order sequence of nodes, which is an array of node values sorted in ascending order. The values in the nodes are unique, allowing us to talk about the kth smallest value. Since the value k is 1-indexed, the first smallest value corresponds to k=1, the second smallest to k=2, and so on.

Flowchart Walkthrough

For the LeetCode problem 230, Kth Smallest Element in a BST, let's analyze and find the most appropriate algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

Yes: While the problem involves a Binary Search Tree, which is a special kind of graph, we can still consider a BST as a graph where each node is connected by edges representing parent-child relationships.

Is it a tree?

Yes: By definition, a Binary Search Tree is a tree, and the problem explicitly mentions that we are working with a BST.

Does the problem involve connectivity?

No: We are not concerned with connectivity between the nodes; instead, we are interested in the property of the nodes themselves, specifically their numerical values and order.

After confirming it's a tree but not involving connectivity issues, the remaining clear pathway for a tree-based problem is to utilize the Depth-First Search algorithm. Depth First Search (DFS) is well suited for tree traversals, such as pre-order, in-order, or post-order. In this specific problem, the goal is to find the kth smallest element, which fits perfectly with an in-order traversal of the BST, a direct application of DFS that visits nodes in the increasing order of their values.

Conclusion: The flowchart leads us to choose DFS for an efficient in-order traversal of the BST to find the kth smallest element.

Intuition

The intuition behind the solution is hinged on the in-order traversal of a BST, which yields the nodes in ascending order. An in-order traversal involves visiting the left subtree, the node itself, and then the right subtree. When this is done for all nodes, we end up visiting nodes in increasing order due to the properties of the BST.

The stated problem is a classic case for using this traversal strategy. However, instead of doing a complete in-order traversal and then finding the kth smallest element, we aim to optimize the process by stopping as soon as we've found the kth smallest element.

The solution uses an iterative approach with a stack to simulate the in-order traversal rather than doing it recursively. We do this by traversing to the leftmost node while pushing all the nodes on the path onto the stack. Once we reach the leftmost node, we start popping nodes from the stack.

Each pop operation corresponds to visiting a node in the in-order sequence. We decrement the value of k each time a node is popped. When k becomes zero, it means we have reached the kth smallest element, and we can stop the traversal and return the value of the current node.

The benefit of this approach is that we only visit k nodes, making the time complexity proportional to k, which could be substantially less than the total number of nodes in the BST.

Learn more about Tree, Depth-First Search, Binary Search Tree and Binary Tree patterns.

Solution Approach

The solution provided uses an iterative approach to perform the in-order traversal of the binary search tree. Here is how the algorithm is implemented:

A stack, denoted by stk, is initialized. It's a fundamental data structure used here to simulate the recursive nature of in-order traversal without actually using recursion. The stack keeps track of nodes along the path that we've yet to visit.
The main loop of the algorithm continues as long as there are nodes to be processed, which is indicated by the conditions while root or stk.
Within the loop, if the current node root is not None, the algorithm pushes this node onto the stack using stk.append(root), and then moves left down the tree by setting root = root.left. This is essentially diving into the leftmost node, adhering to the in-order traversal pattern.
If the current node root is None, which means there is no left child, the algorithm then pops a node from the stack with root = stk.pop(). Popping from the stack is the equivalent of "visiting" a node in the in-order sequence since it has no left children or all its left children have been visited.
After popping a node, k is decremented by 1 to indicate that we have found one more element in the in-order traversal.
When k becomes 0, the loop has reached the kth smallest node, and the value of this node is returned with return root.val. At this point, the traversal stops and does not continue to process the remaining nodes.
If k has not reached 0, it means the kth smallest element has not been found yet, and the algorithm needs to move to the right child of the current node by setting root = root.right. And then the loop will continue to the next iteration to check if there's a left subtree to process from this new node.

The implementation uses a stack to store nodes and a while loop to maintain state, which avoids the overhead of function calls in a recursion, especially if the tree is skewed or the k value is large. With the use of this stack and loop-based approach, the algorithm efficiently realizes the in-order traversal and effectively returns the kth smallest element with the desired complexity.

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Example Walkthrough

Let us consider the following binary search tree (BST) for an example walkthrough:

Suppose we want to find the kth smallest element with k = 3. The in-order traversal of this BST will give us the nodes in ascending order: 1, 2, 3, 4, 5, 6.

We initialize an empty stack stk.
Starting with root as the root of the BST (node with value 5), we follow the left children by moving down to nodes with values 3, then 2, and finally to 1. Each node is pushed onto the stack stk before moving to its left child.

Stack stk status: [5, 3, 2, 1] (top to bottom).
Since node 1 has no left child, we start popping from the stack.
Node 1 is popped; k is decremented to 2.

Stack stk status: [5, 3, 2].
Since node 1 has no right child, we continue popping.
Node 2 is popped; k is decremented to 1.

Stack stk status: [5, 3].
Now, the current root becomes 3 (after popping 2) and hence, we check its right child. Since it has one, we push it onto the stack and set root to node 3's right child.

Stack stk status: [5, 3, 4].
Node 4 has no left child, so we pop it from the stack.
Node 4 is popped; k is decremented to 0.

Stack stk status: [5, 3].
When k is 0, we've found our kth smallest element, which is the value of node 4.

Thus, the value of the 3rd smallest element in the BST is 4. The algorithm successfully finds the kth smallest element using the iterative in-order traversal, with the help of the stack data structure, and stops as soon as it finds the required element without traversing the whole tree.

Solution Implementation

1# Definition for a binary tree node.
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
10        # Initialize an empty list to be used as a stack 
11        stack = []
12
13        # Continue the loop as long as there are nodes to be processed
14        # either in the stack or in the tree starting with the root.
15        while root or stack:
16            # Go to the leftmost node
17            if root:
18                # Push the current node onto the stack
19                stack.append(root)
20                # Move to the left child of the current node
21                root = root.left
22            else:
23                # If there is no left child, pop the node from the stack,
24                # this node is the next in the in-order traversal
25                root = stack.pop()
26                # Decrement k, as we have found one more small element
27                k -= 1
28                # If k is 0, we have found the kth smallest element
29                if k == 0:
30                    # Return the value of kth smallest node
31                    return root.val
32                # Move to the right child of the current node
33                root = root.right
34
35        # The function should always return before reaching this point
36

1class Solution {
2    // Method to find the k-th smallest element in a BST
3    public int kthSmallest(TreeNode root, int k) {
4        // Stack to simulate the in-order traversal iteratively
5        Deque<TreeNode> stack = new ArrayDeque<>();
6      
7        // Continue until we have no unfinished nodes to process
8        while (root != null || !stack.isEmpty()) {
9            // Reach the leftmost node of the current subtree
10            while (root != null) {
11                stack.push(root);
12                root = root.left;
13            }
14            // Process leftmost node
15            root = stack.pop();
16            // Decrement k and if it becomes 0, it means we found our k-th smallest
17            if (--k == 0) {
18                return root.val;
19            }
20            // Move to right subtree to continue the process
21            root = root.right;
22        }
23        // If k is not valid, return 0 as a default value
24        return 0;
25    }
26}
27

1/**
2 * Definition for a binary tree node.
3 */
4struct TreeNode {
5    int val;
6    TreeNode *left;
7    TreeNode *right;
8
9    TreeNode() : val(0), left(nullptr), right(nullptr) {}
10
11    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
12
13    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
14};
15
16class Solution {
17public:
18    /**
19     * Returns the k-th smallest element in a BST.
20     * This is achieved by performing an in-order traversal iteratively.
21     * In-order traversal of a BST yields the nodes in increasing order.
22     * 
23     * @param root A pointer to the root node of the BST.
24     * @param k The index (1-indexed) of the smallest element to be found.
25     * @return The value of the k-th smallest node.
26     */
27    int kthSmallest(TreeNode* root, int k) {
28        // Initialize an empty stack that will be used to perform the in-order traversal.
29        stack<TreeNode*> stk;
30
31        // Continue traversing the tree until we have visited all nodes.
32        while (root || !stk.empty()) {
33            // Go left as long as there is a left child.
34            if (root) {
35                stk.push(root);      // Push the current node onto the stack before moving to left child.
36                root = root->left;   // Move left.
37            } else {
38                // Process the nodes that do not have a left child anymore.
39                root = stk.top();
40                stk.pop();   // Remove the node from the stack.
41                if (--k == 0) {
42                    // If we have reached the k-th smallest, return its value.
43                    return root->val;
44                }
45                // Move to the right child, which will be processed after all its left children.
46                root = root->right;
47            }
48        }
49        // If we are here, it means the k-th smallest element could not be found, return 0.
50        // Although in valid BSTs with at least k nodes, the function will never reach here.
51        return 0;
52    }
53};
54

1// Definition for a binary tree node.
2class TreeNode {
3  val: number;
4  left: TreeNode | null;
5  right: TreeNode | null;
6
7  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
8    this.val = val === undefined ? 0 : val;
9    this.left = left === undefined ? null : left;
10    this.right = right === undefined ? null : right;
11  }
12}
13
14/**
15 * Finds the k-th smallest element in a BST (Binary Search Tree).
16 * @param {TreeNode | null} root - The root of the BST.
17 * @param {number} k - The k-th position to find the smallest element.
18 * @returns {number} - The k-th smallest element in the BST.
19 */
20function kthSmallest(root: TreeNode | null, k: number): number {
21  // Decrement k-th position, initially the number of elements to skip
22  let elementsToSkip: number = k;
23
24  /**
25   * In-order traversal of the BST that keeps track of the traversal sequence
26   * and searches for the k-th smallest element.
27   * @param {TreeNode | null} node - The current node in the traversal.
28   * @returns {number} - The value of the k-th smallest element or -1 if not found.
29   */
30  function inOrderTraversal(node: TreeNode | null): number {
31    // Base case: if the current node is null, return -1 to indicate no further nodes in this path
32    if (node === null) {
33      return -1;
34    }
35
36    // Recurse on the left subtree
37    const leftValue = inOrderTraversal(node.left);
38    // If a valid non-negative value is found from the left subtree, return it as the result
39    if (leftValue !== -1) {
40      return leftValue;
41    }
42
43    // Decrement elementsToSkip to move towards the k-th smallest element
44    elementsToSkip--;
45    // If elementsToSkip becomes 0, current node's value is the k-th smallest, return it
46    if (elementsToSkip === 0) {
47      return node.val;
48    }
49
50    // Recurse on the right subtree
51    return inOrderTraversal(node.right);
52  }
53
54  // Perform in-order traversal and return value of the k-th smallest element
55  return inOrderTraversal(root);
56}
57
58// Note: This algorithm assumes the BST property where for every node, all values in its left subtree
59// are smaller than its value, and all values in the right subtree are greater. The in-order traversal
60// then guarantees that the elements are visited in ascending order.
61

Time and Space Complexity

The code implements an in-order traversal of a binary search tree to find the kth smallest element. Let's analyze the time complexity and space complexity:

Time Complexity

The time complexity of the code is O(H + k), where H is the height of the tree, and k is the input parameter. This is because in the worst case, the code performs the in-order traversal to the left-most node (which could be a height of H), and then continues to traverse the tree until the kth smallest element is found. However, since this is a binary search tree, in the average case where the tree is balanced, the height H can be considered as log(N) and thus the average time complexity would be O(log(N) + k).

Space Complexity

The space complexity is determined by the size of the stack stk which, in the worst case, will hold all the nodes of one branch of the tree at the same time. In the worst case, this can be O(H) - which is the height of the tree if the tree is skewed. However, in an average case where the tree is reasonably balanced, the height H would be log(N) and hence we can consider the space complexity as O(log(N)).

Learn more about how to find time and space complexity quickly using problem constraints.

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