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);
     }
 }