leetcode解题: Kth Smallest Element in a BST (230)

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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： In-order traversal O（N）， N is number of elements

这题可以看成是一个in-order traversal的直接应用，因为BST的in-order traversal是一个有序数组，那么我们就可以根据这个性质，对数进行遍历直到找到第k个node。
C++

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/**
 * 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) {
        // implement a in-order-traversal
        int count = 0;
        int res;
        stack<TreeNode*> s;
        TreeNode* cur = root;
        while (cur || !s.empty()) {
            if (cur) {
                s.push(cur);
                cur = cur->left;
            } else {
                ++count;
                TreeNode* temp = s.top();
                s.pop();
                if (count == k) {
                    res = temp->val;
                    break;
                }
                cur = temp->right;
            }
        }
        return res;
    }
};

Java

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        // in-order traversal iterative solution
        Stack<TreeNode> stack = new Stack<TreeNode>();        
        TreeNode current = root;
        int count = 0;
        int val = 0;
        while ( current != null || !stack.isEmpty()) {
            if (current != null) {
                stack.push(current);
                current = current.left;                
            } else {
                TreeNode temp = stack.pop();
                val = temp.val;
                count++;
                current = temp.right;
                if (count == k) {
                    return val;
                }
            }
        }
        return val;
    }
}

Follow up : O(logN)

参考这篇文章的思路
如果可以修改每一个tree的node的结构，而使其保存leftcount和totalcount。那么一开始建立树需要O(N)的时间，而之后每一次insert和查找则只需要O(logN)的时间。

lang: java
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class Solution {
    class SuperNode {
        int val;
        int count;
        int total;
        SuperNode left;
        SuperNode right;
        public SuperNode(int val) {
            this.val = val;
            count = 0;
            total = 1;
            left = null;
            right = null;
        }
    };
    public SuperNode build(TreeNode root) {
        if (root == null) return null;
        SuperNode superRoot = new SuperNode(root.val);
        SuperNode left = build(root.left);
        SuperNode right = build(root.right);
        superRoot.left = left;
        superRoot.right = right;
        if (left != null) {
            superRoot.count = left.total;
        } else {
            superRoot.count = 0;
        }
        superRoot.total += ((left == null ? 0 : left.total) + (right == null ? 0 : right.total));
        return superRoot;
    }
    public int kthSmallest(TreeNode root, int k) {
        SuperNode superRoot = build(root);
        return kth(superRoot, k);
    }
    private int kth(SuperNode root, int k) {
        if (root == null || k <= 0) {
            return 0;
        }        
        if (k == root.count + 1) {
            return root.val;
        } else if (k < root.count + 1) {
            return kth(root.left, k);
        } else {
            return kth(root.right, k - root.count - 1);
        }
    }
}