# Second Place - Amazon Top Interview Questions

### Problem Statement :

```Given a binary tree root, return the depth of the second deepest leaf. Note that if there are multiple deepest leaves, the second deepest leaf is the next highest one.

The root has a depth of 0 and you can assume the answer is guaranteed to exist for the trees given.

Constraints

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

Example 1

Input

root = [1, [2, null, null], [3, [4, [6, null, null], null], [5, null, [7, null, null]]]]

Output

1

Explanation

The the second deepest leaf is 2 which exists at depth 1```

### Solution :

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

int solve(Tree* root) {
queue<Tree*> q;
q.push(root);

int depth = 0;
int deepest = 0;
int res = 0;
while (!q.empty()) {
int n = q.size();
while (n--) {
Tree* t = q.front();
q.pop();

if (t->left == t->right) {
if (depth > deepest) {
res = deepest;
deepest = depth;
} else if (depth > res and depth < deepest)
res = depth;
}

if (t->left) q.push(t->left);
if (t->right) q.push(t->right);
}
depth++;
}

return res;
}```
```

```                        ```Solution in Java :

import java.util.*;

/**
* public class Tree {
*   int val;
*   Tree left;
*   Tree right;
* }
*/
class Solution {
int max1 = -1, max2 = -1;

public void dfs(Tree root, int depth) {
if (root == null)
return;
this.dfs(root.left, depth + 1);
this.dfs(root.right, depth + 1);
if (root.left == null && root.right == null)
if (depth == max1)
return;
else if (depth > max1) {
max2 = max1;
max1 = depth;
} else if (depth > max2)
max2 = depth;
}

public int solve(Tree root) {
dfs(root, 0);
return max2;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, root):
depths = set()

def dfs(node, d=0):
if node:
if not node.left and not node.right:
dfs(node.left, d + 1)
dfs(node.right, d + 1)

dfs(root)
return max(depths)```
```

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

## Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t