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.Show More Hint

Credits:

Special thanks to @ts for adding this problem and creating all test cases.

两种思路：

1.空间换时间

BST的特性是，如果按照中序排列，得到的递增序；所以可以使用一个stack进行中序遍历，直到找到第K个元素；

2. 树树的结点数

对于每个节点root，计算以它为根节点的树的节点数，计作S。

如果S==K，返回root->val;

如果S > K，在root的左子树里面查找第K小元素；

如果S

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Stack<TreeNode> s = new Stack<>();
        TreeNode p=root;
        s.push(p);
        int cnt=0;
        do{
            while(p!=null){
                s.push(p);
                p=p.left;
            }
            TreeNode top=s.pop();
            cnt++;
            if(cnt==k){
                return top.val;
            }
            p=top.right;
        }while(!s.empty());
        return 0;
    }
}

思路2代码：

/**
 * 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) {
        int left=findNodesSum(root->left);
        if(left+1==k){
            return root->val;
        }else if(left+1<k){
            return kthSmallest(root->right,k-left-1);
        }else{
            return kthSmallest(root->left,k);
        }
    }
    int findNodesSum(TreeNode* root){
        if(!root){
            return 0;
        }
        int sum = findNodesSum(root->left)+findNodesSum(root->right)+1;
        return sum;
    }
};

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