2016年2月26日 星期五

LEET code -- 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:

  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).

就是找出DFS他parse的順序找出來

為了可以提早結束  我用-1代表找到
rank代表現在找到第幾個數字了


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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) {
        if(root==NULL)return 0;
        rank=0;
        DFS(root,k);
        return reuslt;
        
    }
    int DFS(TreeNode* root,int level){
        if(root==NULL)return rank;
        int left=DFS(root->left,level);
        if(left<0)return -1;
        rank=rank+1;
        if(level==rank){
            reuslt=root->val;
            return -1;
        }
        int right=DFS(root->right,level);
        if(right <0)return -1;
        return 1;
    }
private:
    int rank;
    int reuslt;
};

--------------------------


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class Solution {
public:
    TreeNode* current = NULL;
    stack<TreeNode*> s;
    int kthSmallest(TreeNode* root, int k) {
        current = root;
        int count = 1;
        while(1)
        {
        while(current)
        {
            s.push(current);
            current = current->left;
        }
        if(count == k)
        {
            TreeNode* node = s.top();
            return node->val;
        }
        if(count != k)
        {
            TreeNode* node = s.top();
            s.pop();
            current = node->right;
            count++;
        }
        }
    }
};


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