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:

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：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;
}
}