Kth Smallest Element in a BST —— Leetcode

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:

  1. Try to utilize the property of a BST.
  2. What if you could modify the BST node's structure?
  3. The optimal runtime complexity is O(height of BST).
用的BST性质就是,左侧都比root小,右侧都比root大。

如果每次用递归来计算左子树或右子树的结点个数,类似于二分法,但每次计算左子树节点个数会有很多重复开销;如果按照提示,改变TreeNode的存储结构,再添加一个leftCount属性,那么就比较容易计算出结果了。

下面提供了一种用stack来保存节点的方法,思想是将节点入栈,保证从最小节点开始遍历直到找到第k个,时间复杂度是O(k);

/**
 * 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) {
        stack<TreeNode *> st;
        TreeNode* p = root;
        while(p || !st.empty()) {
            while(p) {
                st.push(p);
                p = p->left;
            }
            p = st.top();
            if(--k == 0) {
                return p->val;
            }
            
            st.pop();
            p = p->right;
        }
    }
};



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