Kth Smallest Element in a BST 解答

Question

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).

Solution 1 -- Inorder Traversal

Again, we use the feature of inorder traversal of BST. But this solution is not best for follow up. Time complexity O(n), n is the number of nodes.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        TreeNode current = root;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        while (current != null || !stack.empty()) {
            if (current != null) {
                stack.push(current);
                current = current.left;
            } else {
                TreeNode tmp = stack.pop();
                k--;
                if (k == 0) {
                    return tmp.val;
                }
                current = tmp.right;
            }
        }
        return -1;
    }
}

Solution 2 -- Augmented Tree

The idea is to maintain rank of each node. We can keep track of elements in a subtree of any node while building the tree. Since we need K-th smallest element, we can maintain number of elements of left subtree in every node.

Assume that the root is having N nodes in its left subtree. If K = N + 1, root is K-th node. If K < N, we will continue our search (recursion) for the Kth smallest element in the left subtree of root. If K > N + 1, we continue our search in the right subtree for the (K – N – 1)-th smallest element. Note that we need the count of elements in left subtree only.

Time complexity: O(h) where h is height of tree.

(referrence: GeeksforGeeks)

Here, we construct tree in a way that is taught during Algorithm class.

"size" is an attribute which indicates number of nodes in sub-tree rooted in that node.

Time complexity: constructing tree O(n), find Kth smallest number O(h).

start:
if K = root.leftElement + 1
   root node is the K th node.
   goto stop
else if K > root.leftElements
   K = K - (root.leftElements + 1)
   root = root.right
   goto start
else
   root = root.left
   goto srart

stop

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class ImprovedTreeNode {
    int val;
    int size; // number of nodes in the subtree that rooted in this node
    ImprovedTreeNode left;
    ImprovedTreeNode right;
    public ImprovedTreeNode(int value) {val = value;}
}

public class Solution {
    
    // Construct ImprovedTree recursively
    public ImprovedTreeNode createAugmentedBST(TreeNode root) {
        if (root == null)
            return null;
        ImprovedTreeNode newHead = new ImprovedTreeNode(root.val);
        ImprovedTreeNode left = createAugmentedBST(root.left);
        ImprovedTreeNode right = createAugmentedBST(root.right);
        newHead.size = 1;
        if (left != null)
            newHead.size += left.size;
        if (right != null)
            newHead.size += right.size;
        newHead.left = left;
        newHead.right = right;
        return newHead;
    }
    
    public int findKthSmallest(ImprovedTreeNode root, int k) {
        if (root == null)
            return -1;
        ImprovedTreeNode tmp = root;
        int leftSize = 0;
        if (tmp.left != null)
            leftSize = tmp.left.size;
        if (leftSize + 1 == k)
            return root.val;
        else if (leftSize + 1 > k)
            return findKthSmallest(root.left, k);
        else
            return findKthSmallest(root.right, k - leftSize - 1);
    }
    
    public int kthSmallest(TreeNode root, int k) {
        if (root == null)
            return -1;
        ImprovedTreeNode newRoot = createAugmentedBST(root);
        return findKthSmallest(newRoot, k);
    }
}