Construct BST from given preorder traversal | Set 3 (Naive Method)

Given preorder traversal of a binary search tree, construct the BST.

For example, if the given traversal is {10, 5, 1, 7, 40, 50}, then the output should be root of following tree.

    10
   /   \
  5     40
 /  \      \
1    7      50  

We have discussed methods to construct binary search tree in previous posts.Here is another method to construct binary search tree when given preorder traversal.

We know that the inorder traversal of the BST gives the element in non-decreasing manner. Hence we can sort the given preorder traversal to obtain the inorder traversal of the binary search tree.

We have already learnt the method to construct tree when given preorder and inorder traversals in this post. We will now use the same method to construct the BST.



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#include <bits/stdc++.h>
using namespace std;
  
// A BST node has data, pointer to left 
// child and pointer to right child
struct Node {
    int data;
    Node *left, *right;
};
  
// A utility function to create new node
Node* getNode(int data)
{
    Node* temp = new Node();
    temp->data = data;
    temp->left = temp->right = NULL;
  
    return temp;
}
  
/* Recursive function to construct BST
Inorder traversal in[] and Preorder traversal
pre[]. Initial values of inStart and inEnd should be
0 and n -1.*/
Node* buildBTRec(int in[], int pre[], int inStart, 
            int inEnd, unordered_map<int, int>& m)
{
    static int preIdx = 0;
    if (inStart > inEnd)
        return NULL;
  
    // Pick current node from Preorder traversal
    // using preIndex and increment preIndex
    int curr = pre[preIdx];
    ++preIdx;
  
    Node* temp = getNode(curr);
  
    // If this node has no children then return
    if (inStart == inEnd)
        return temp;
  
    // Else find the index of this node in
    // inorder traversal
    int idx = m[curr];
  
    // Using this index construct left and right subtrees
    temp->left = buildBTRec(in, pre, inStart, idx - 1, m);
    temp->right = buildBTRec(in, pre, idx + 1, inEnd, m);
  
    return temp;
}
  
// This function mainly creates a map to store 
// the indices of all items so we can quickly 
// access them later.
Node* buildBST(int pre[], int n)
{
    // Copy pre[] to in[] and sort it
    int in[n];
    for (int i = 0; i < n; i++)
        in[i] = pre[i];
    sort(in, in + n);
  
    unordered_map<int, int> m;
    for (int i = 0; i < n; i++)
        m[in[i]] = i;
    return buildBTRec(in, pre, 0, n - 1, m);
}
  
// Utility function to do Inorder traversal of the tree
void inorderTraversal(Node* root)
{
    if (root == NULL)
        return;
    inorderTraversal(root->left);
    cout << root->data << " ";
    inorderTraversal(root->right);
}
  
// Driver Program
int main()
{
    int pre[] = { 100, 20, 10, 30, 200, 150, 300 };
    int n = sizeof(pre) / sizeof(pre[0]);
  
    Node* root = buildBST(pre, n);
  
    // Let's test the built tree by printing its 
    // Inorder traversal
    cout << "Inorder traversal of the tree is \n";
    inorderTraversal(root);
    return 0;
}

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Output:

Inorder traversal of the tree is 
10 20 30 100 150 200 300

Time Complexity: Sorting takes O(nlogn) time for sorting and constructing using preorder and inorder traversals takes linear time. Hence overall time complexity of the above solution is O(nlogn).

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