Given a binary search tree, write a function kthSmallest to find thekth 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?
在二叉搜索树种，找到第K个元素。
算法例如以下：
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。
代码例如以下：
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);
}
}
};