Leetcode|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?

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).

解法1：中序遍历

int kthSmallest(TreeNode* root, int k) {
if(!root) return 0;
stack<TreeNode*> s;
while(root||s.size()>0){
if(root){
s.push(root);
root=root->left;
}
else{
TreeNode *p=s.top();
s.pop();
k--;
if(k==0) return p->val;
root=p->right;
}
}
}

解法2：构造新的BST节点，包含左子树的节点个数信息。如果root节点的count值为k-1，那么就是说有k-1个点比他的值小，该节点就是所求点。如果该节点count值小于k-1（假设为N），寻找他的右孩子，相当于寻找root->right,的count值为（k-1-N). 如果root节点的count值比k-1大，那么继续搜索root->left直到count值为k-1;