Tree From Pre/Inorder Traversals - Amazon Top Interview Questions


Problem Statement :


Given a list of unique integers preorder and another list of unique integers inorder, representing the pre-order and in-order traversals of a binary tree, reconstruct the tree and return the root.

Constraints

n ≤ 100,000 where n is the length of preorder and inorder

Example 1

Input

preorder = [4, 2, 1, 0, 3]

inorder = [2, 1, 0, 3, 4]

Output

[4, [2, null, [1, null, [0, null, [3, null, null]]]], null]


Solution :



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                        Solution in C++ :

Tree* make_tree(vector<int>& preorder, int idx, vector<int>& inorder, int l, int r) {
    if (idx >= preorder.size() || l >= r) return nullptr;
    Tree* root = new Tree();
    root->val = preorder[idx];
    auto it = find(inorder.begin() + l, inorder.begin() + r, root->val);
    int mid = it - inorder.begin();
    root->left = make_tree(preorder, idx + 1, inorder, l, mid);
    root->right = make_tree(preorder, idx + mid - l + 1, inorder, mid + 1, r);
    return root;
}

Tree* solve(vector<int>& preorder, vector<int>& inorder) {
    return make_tree(preorder, 0, inorder, 0, inorder.size());
}
                    


                        Solution in Java :

import java.util.*;

/**
 * public class Tree {
 *   int val;
 *   Tree left;
 *   Tree right;
 * }
 */
class Solution {
    int rootCount = 0;
    public Tree solve(int[] preorder, int[] inorder) {
        if (preorder.length == 0 || inorder.length == 0)
            return null;

        int[] left = null;
        int[] right = null;

        int root = preorder[rootCount];
        rootCount++;
        // System.out.println("Root is "+ root);
        // System.out.println("Inorder tree is " + Arrays.toString(inorder));
        for (int i = 0; i < inorder.length; i++) {
            int num = inorder[i];
            if (num == root) {
                left = Arrays.copyOfRange(inorder, 0, i);
                right = Arrays.copyOfRange(inorder, i + 1, inorder.length);
                break;
            }
        }
        Tree tree = new Tree(root);
        // System.out.println("Left Inorder tree is " + Arrays.toString(left));
        // System.out.println("Right Inorder tree is " + Arrays.toString(right));

        tree.left = solve(preorder, left);
        tree.right = solve(preorder, right);

        return tree;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, preorder, inorder):
        if inorder:
            i = inorder.index(preorder.pop(0))
            return Tree(
                inorder[i],
                left=self.solve(preorder, inorder[:i]),
                right=self.solve(preorder, inorder[i + 1 :]),
            )


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