230. Kth Smallest Element in a BST

最新推荐文章于 2021-04-09 14:18:22 发布

原创最新推荐文章于 2021-04-09 14:18:22 发布 · 276 阅读

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#traverse tree

本文介绍了一种高效的方法来找到二叉搜索树中第K小的元素，通过利用二叉搜索树的性质进行中序遍历，并记录经过的节点数量，从而在O(height of BST)的时间复杂度内完成任务。

Problem

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).

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

Solution

in order 遍历二叉树，

如果节点为空就直接返回，否则先遍历完左边，更新走过的个数 K，之后比较K是否想要的。

/**
 * 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 {
    void helper( TreeNode* root, int& k, int& rst){
        if(!root) return;
        helper(root->left, k, rst);
        k--;
        
        if( k == 0){
            rst = root->val;
            return;
        }
        
        helper(root->right, k, rst);
    }
public:
    int kthSmallest(TreeNode* root, int k) {
        int rst;
        helper(root, k, rst);
        return rst;
    }
};

Follow - up

那就给每个树节点增加一个值：左边节点的个数（numLeftCount）.

if ( k == numLeftCount + 1) return;

if(k < numLeftCount + 1)

go left;

else

k -= (numLeftCount + 1);

go right;