230. Kth Smallest Element in a BST

最新推荐文章于 2024-01-19 11:14:28 发布

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最新推荐文章于 2024-01-19 11:14:28 发布

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分类专栏： leetcode刷题

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本文介绍了一个解决寻找二叉搜索树中第K小元素的问题的方法，并提供了一种递归解决方案。此外，还讨论了如何在频繁修改二叉搜索树的情况下优化查找过程。

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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?

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
   void __kthSmallest(TreeNode* root, int k, int& ret, TreeNode** result){
	 if(!root){
		 return;
	 }

	 if(root->left){
		 __kthSmallest(root->left, k, ret, result);
	 }

	 ret++;
	 if(ret == k){
		 *result = root;
		 return;
	 }

	 if(root->right){
		 __kthSmallest(root->right, k, ret, result);
	 }
 }
 int kthSmallest(TreeNode* root, int k) {
	 TreeNode* result = new TreeNode(-1);
	 int ret = 0;
	 __kthSmallest(root, k, ret, &result);
	 return result->val;
 }
};

int kthSmallest(TreeNode* root, int& k) {
    if (root) {
        int x = kthSmallest(root->left, k);
        return !k ? x : !--k ? root->val : kthSmallest(root->right, k);
    }
}