# Inorder Traversal - Amazon Top Interview Questions

### Problem Statement :

```Given a binary tree root, return an inorder traversal of root as a list.

Bonus: Can you do this iteratively?

Constraints

n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

root = [1, null, [159, [80, null, null], null]]

Output

[1, 80, 159]```

### Solution :

```                        ```Solution in C++ :

vector<int> solve(Tree* root) {  // Morris traversal - Time: O(N), Space: O(1)
vector<int> inorder;

while (root) {
if (root->left) {
Tree* predecessor = root->left;
while (predecessor->right && predecessor->right != root)
predecessor = predecessor->right;

if (predecessor->right) {
inorder.push_back(root->val);
root = root->right;
predecessor->right = nullptr;
} else {
predecessor->right = root;
root = root->left;
}
} else {
inorder.push_back(root->val);
root = root->right;
}
}

return inorder;
}```
```

```                        ```Solution in Java :

import java.util.*;

/**
* public class Tree {
*   int val;
*   Tree left;
*   Tree right;
* }
*/
class Solution {
public int[] solve(Tree root) {
List<Integer> list = new ArrayList<>();
list = inOrder(root, new ArrayList<>());
int[] arr = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
arr[i] = list.get(i);
}
return arr;
}
public List<Integer> inOrder(Tree root, List<Integer> list) {
if (root == null)
return list;
Stack<Tree> st = new Stack<>();
Tree temp = root;
while (temp != null || st.size() > 0) {
while (temp != null) {
st.push(temp);
temp = temp.left;
}

temp = st.pop();
temp = temp.right;
}
return list;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, root):
if not root:
return []
# iteratively
stack = [root.right, root.val, root.left]
traversal = []

while stack:
currNode = stack.pop()
if type(currNode) == int:
traversal.append(currNode)
elif currNode is None:
continue
else:
stack.extend([currNode.right, currNode.val, currNode.left])

return traversal```
```

## Square-Ten Tree

The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the

## Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

## Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

## Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

## Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

## Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ