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); } } };