【LeetCode】230. Kth Smallest Element in a BST

二叉搜索树第K小元素

最新推荐文章于 2023-06-03 15:29:39 发布

原创最新推荐文章于 2023-06-03 15:29:39 发布 · 287 阅读

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本文介绍了一种寻找二叉搜索树中第K小元素的方法，通过中序遍历实现，并探讨了如何优化算法以应对频繁修改的场景。

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).

class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        
        stack<TreeNode*> stack;
        TreeNode * p=root;
        int result;
        int count=k;
        
        while((!stack.empty() || p!=NULL)){
            
            if(p!=NULL){
                stack.push(p);
                p=p->left;
            }else{
                
                count--;
                if(count==0){
                    break;
                }

                p=stack.top();
                stack.pop();
                p=p->right;
                
            }
        }
        
        result =stack.top()->val;
        
        return result;
    }
};

思路心得：

这题我的做法其实就是为了中序遍历加了一个计数器。然后就能记录下来BST中的元素大小了。

但是这题的思考部分是要为结点添加子树的大小这个数据。

感觉和非递归遍历BST比谁快?

虽然如果事先就有子树的大小的话确实是O(树高)