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 :
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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