Build a binary tree from a given preorder traversal

Question

I have a preorder traversal of a binary tree that is stored in an array and I would like to recreate the binary tree based on this traversal. My array looks like this: {NNNLLNLLNLNLNNLLNLL}, where N represents a node and L represents a leaf. I would like to do this recursively but I am having trouble coming up with an algorithm. Any suggestions would be very appreciated.

Answer 1

This should work assuming every node has 2 or 0 descendants (a tree that satisfies this property is called full or strict binary tree )

void create_from_traversal(Node* root, int& index) {
    if (traversal[index] == 'L') {
        root->left = root->right = NULL;
        return;
    }
    root->left = new Node();
    create_from_traversal(root->left, ++index);
    root->right = new Node();
    create_from_traversal(root->right, ++index);
}

Complete example with check:

#include <string>
#include <iostream>

class Node {
public:
    Node* left;
    Node* right;
};

std::string traversal = "NNNLLNLLNLNLNNLLNLL";

void create_from_traversal(Node* root, int& index) {
    if (traversal[index] == 'L') {
        root->left = root->right = NULL;
        return;
    }
    root->left = new Node();
    create_from_traversal(root->left, ++index);
    root->right = new Node();
    create_from_traversal(root->right, ++index);
}

void print_traversal(Node* root) {
    if (root->left == NULL) {
        std::cout << "L";
        return;
    }
    std::cout << "N";
    print_traversal(root->left);
    print_traversal(root->right);
}

int main() {
    Node* root = new Node();
    int index = 0;
    create_from_traversal(root, index);

    // Does it work?
    print_traversal(root); // Output should be equal to given traversal
    std::cout << std::endl;
}

Output:

NNNLLNLLNLNLNNLLNLL

Answer 2

You need one more traversal before you can reconstruct the tree. Given any two traversals among the three (Pre, Post, In) you can reconstruct. But given only one it is not possible to uniquely reconstruct the tree.