Construct bst from preorder

#1

code: https://ide.codingblocks.com/s/584574

problem: https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/

#2

I have put a check for NULL in line 30 but still it is showing an error due to left being NULL.

#3

you just have to do this

TreeNode* build(TreeNode* &root,int val){
        if(root == nullptr){
            //cout<<"Now val "<<val<<endl;
            return new TreeNode(val);;
        }
        //cout<<root->val<<" ";
        if(root->val < val) root->right = build(root->right,val);
        if(root->val > val) root->left = build(root->left,val);
        return root;
    }
    TreeNode* bstFromPreorder(vector<int>& preorder) {
        if(preorder.size()==0) return nullptr;
        TreeNode* root = nullptr;
        for(int i=0;i<preorder.size();i++){
            root = build(root,preorder[i]);
        }
        return root;
    }

what’s the approach you are using.

#4


second method of this blog

#5

both have same time complexity, you can go with the method i have mentioned above.

#6

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