leetcode 230: Kth Smallest Element in a BST

最新推荐文章于 2024-06-07 05:21:43 发布

原创最新推荐文章于 2024-06-07 05:21:43 发布 · 264 阅读

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本文介绍了一种高效查找二叉搜索树中第K小元素的方法。通过改进的中序遍历，该算法能在找到目标元素时提前终止遍历过程，从而提升效率。

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Use the inorder traversal and get all numbers in the BST in order. Then return the kth number.

/**
 * 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:
    int kthSmallest(TreeNode* root, int k) {
        vector<int> nums;
        inorder(root,nums);
        return nums[k-1];
    }
    void inorder(TreeNode* root,vector<int>& nums)
    {
        if(!root)
            return;
        inorder(root->left,nums);
        nums.push_back(root->val);
        inorder(root->right,nums);
    }
};

The following is the updated version with better run time, in which I do not need to traverse the whole tree. When I reach the kth smallest number, I just return that number to the top.

/**
 * 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:
    int kthSmallest(TreeNode* root, int k) {
        return find(root,k);
    }
    int find(TreeNode* root,int &k)
    {
        if(!root)
            return 0;
        int l=find(root->left,k);
        if(k==0)//the kth number is in the left subtree
            return l;
        k--;
        if(k==0)//the kth number is this root
            return root->val;
        return find(root->right,k);//the kth number is in the right subtree
    }
};