LeetCode题解:Kth Smallest Element in a BST

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:

Try to utilize the property of a BST.
What if you could modify the BST node’s structure?
The optimal runtime complexity is O(height of BST).

题意:给定二叉搜索树,找出第K小的数。假如二叉树经常会被修改,而且频繁需要查找第K小的数呢?

解题思路:DFS

代码:

/**
 * 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) {
        if(root == null){
            return 0;
        }

        int leftNodeNums = countNodeNums(root.left);
        if(leftNodeNums == k - 1){
            return root.val;
        }else if(leftNodeNums < k){
            return kthSmallest(root.right, k - leftNodeNums - 1);
        }else{
            return kthSmallest(root.left, k);
        }
    }

    private int countNodeNums(TreeNode root){
        if(root == null){
            return 0;
        }

        return 1 + countNodeNums(root.left) + countNodeNums(root.right);
    }
}

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