230. Kth Smallest Element in a BST java solutions

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.Show More Hint 

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

 

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    ArrayList<Integer> ans = new ArrayList<Integer>();
    public int kthSmallest(TreeNode root, int k) {
        InorderTree(root);
        return ans.get(k-1);
    }
    
    public void InorderTree(TreeNode node){
        if(node != null){
            InorderTree(node.left);
            ans.add(node.val);
            InorderTree(node.right);
        }
    }
}

非递归算法使用栈：

 

/**
 * 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) {
        Stack<TreeNode> s = new Stack<TreeNode>();
        s.push(root);
        TreeNode node = root;
        while(!s.isEmpty()){
            node = s.peek();
            if(node.left != null){
                s.push(node.left);
                node.left = null;
            }else{
                s.pop();
                k--;
                if(k == 0) return node.val;
                if(node.right != null) s.push(node.right);
            }
        }
        return node.val;
    }
}