[leetcode]230. Kth Smallest Element in a BST

[leetcode]230. Kth Smallest Element in a BST

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Analysis

ummmm—— [每天刷题并不难。。。其实已经很多天没刷了]

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.

Implement

方法一（中序遍历BST，然后输出第k个元素）

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

方法二（Follow up里面提到BST可能频繁的插入删除等，所以可以用分治法，先确定第K小的元素在左子树还是右子树，然后再继续遍历左子树或者右子树）

/**
 * 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) {
        int cnt = count(root->left);
        if(k <= cnt)
            return kthSmallest(root->left, k);
        else if(k > cnt+1)
            return kthSmallest(root->right, k-cnt-1);
        return root->val;
    }
    int count(TreeNode* node){
        if(!node)
            return 0;
        return 1+count(node->left)+count(node->right);
    }
    
};