leetcode——230——Kth Smallest Element in a BST
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?
思路一：
对二叉搜索树中序遍历，第k个数是第K小的元素
/**
* 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 num=0;//记录次数
int re;//记录最终结果
int kthSmallest(TreeNode* root, int k) {
digui(root,k);
return re;
}
void digui(TreeNode* root, int k)
{
if(root->left!=NULL)
{
digui(root->left,k);
}
num++;
if(num==k)
re = root->val;
if(root->right!=NULL)
{
digui(root->right,k);
}
return ;
}
};
思路二：
在二叉搜索树种，找到第K个元素。
算法如下：
1、计算左子树元素个数left。
2、 left+1 = K，则根节点即为第K个元素
left >=k, 则第K个元素在左子树中，
left +1 <k, 则转换为在右子树中，寻找第K-left-1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 calcTreeSize(TreeNode* root){
if (root == NULL)
return 0;
return 1+calcTreeSize(root->left) + calcTreeSize(root->right);
}
int kthSmallest(TreeNode* root, int k) {
if (root == NULL)
return 0;
int leftSize = calcTreeSize(root->left);
if (k == leftSize+1){
return root->val;
}else if (leftSize >= k){
return kthSmallest(root->left,k);
}else{
return kthSmallest(root->right, k-leftSize-1);
}
}
};