leetcode--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:

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).

题意：在BST树中，查找第k小的节点值

分类：二叉树

解法1：BST树很多问题我们都可以使用中序遍历来解决，这个也不例外

我们设置一个当前访问的元素个数cur，在中序遍历过程中，每要访问一个节点，cur加1

当cur和k相等时，这就是我们要找的节点，返回即可

利用递归来进行中序遍历，如果左子树找不到，则判断是不是当前节点，不是，则在右子树接着找

如果左子树找到了，直接返回

/**
 * 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 cur = 0;
    public int kthSmallest(TreeNode root, int k) {
        TreeNode res = inorder(root,k);
        if(res!=null) return res.val;
        return -1;
    }
    
    public TreeNode inorder(TreeNode root,int k){
        if(root==null) return null;
        if(root.left!=null){
            TreeNode left = inorder(root.left,k);//先在左子树找
            if(left!=null) return left;//如果找到了，返回
        }
        cur++;//每次访问节点，都加1
        if(cur==k) return root;//判断是不是当前节点
        if(root.right!=null){//最后在右子树找
            TreeNode right = inorder(root.right,k);
            if(right!=null) return right;
        }
        return null;
    }
}