[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?

 

分析：二叉搜索树中序遍历递增有序。

相关题目：《剑指offer》63

递归版本：

/**
 * 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) {
        TreeNode* p = kthSmallest(root, &k);
        return p->val;
    }
    
    TreeNode* kthSmallest(TreeNode* root, int* k) {
        TreeNode* p = NULL;
        
        if (root->left != NULL) {
            p = kthSmallest(root->left, k);
        }
        
        if (p == NULL) {
            if (*k == 1)
                p = root;
            (*k)--;
        }
        
        if (p == NULL && root->right != NULL)
            p = kthSmallest(root->right, k);
            
        return p;
    }
};

 

 

迭代版本：

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