Day24. ConstructBinarySearchTreeFromPreorderTraversal.cpp

/*

Return the root node of a binary search tree that matches the given preorder traversal.

(Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has a value > node.val.  Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.)

It's guaranteed that for the given test cases there is always possible to find a binary search tree with the given requirements.

Example 1:

Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]

 https://assets.leetcode.com/uploads/2019/03/06/1266.png

Constraints:

1. 1 <= preorder.length <= 100
2. 1 <= preorder[i] <= 10^8
3. The values of preorder are distinct.

*/

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* help(TreeNode* root,int key)
    {
        if(root==NULL) return new TreeNode(key);
        else if(root->val>key) root->left=help(root->left,key);
        else root->right=help(root->right,key);
        
        return root;
    }
    
    
    TreeNode* bstFromPreorder(vector<int>& preorder)
    {
        TreeNode* root=NULL;
        
        for(int i=0;i<preorder.size();i++)
        {
            root=help(root,preorder[i]);
        }
        return root;
    }
};