**Symmetric Binary Tree - Amazon Top Interview Questions**

### Problem Statement :

Given the root to a binary tree root, return whether it is symmetric. Constraints n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [0, [2, [5, null, null], [1, null, null]], [2, [1, null, null], [5, null, null]]] Output True Example 2 Input root = [1, [2, null, null], [3, null, null]] Output False Example 3 Input root = [1, [2, null, null], null] Output False Example 4 Input root = [1, null, [2, null, null]] Output False

### Solution :

` ````
Solution in C++ :
bool solve(Tree* root) {
if (!root) return true;
stack<Tree*> p, q;
p.push(root->left), q.push(root->right);
while (!p.empty() && !q.empty()) {
auto m = p.top(), n = q.top();
p.pop(), q.pop();
if (!m && !n) continue;
else if (!m || !n) return false;
if (m->val != n->val)
return false;
p.push(m->left), p.push(m->right);
q.push(n->right), q.push(n->left);
}
return p.empty() && q.empty();
}
```

` ````
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
public boolean solve(Tree root) {
return versionA(root);
// return versionB(root);
}
// breadth first search, with padding for null child nodes
// performance: 62%
// This also works when Tree is non-bianry, i.e. more than 2 child nodes for a parent node.
boolean versionA(Tree root) {
if (root == null) {
return true;
}
// a special NULL node
Tree padding = new Tree();
ArrayList<Tree> list = new ArrayList<>();
list.add(root);
while (!list.isEmpty()) {
if (!isValid(list, padding)) {
return false;
}
list = getNextLevel(list, padding);
}
return true;
}
// check as if we were checking palindrome: all the nodes at one level
boolean isValid(ArrayList<Tree> list, Tree padding) {
int head = 0;
int tail = list.size() - 1;
while (head < tail) {
if (!isValid(list.get(head), list.get(tail), padding)) {
return false;
}
head++;
tail--;
}
return true;
}
boolean isValid(Tree a, Tree b, Tree padding) {
if ((a == padding) && (b == padding)) {
return true;
}
if ((a == padding) || (b == padding)) {
return false;
}
// no padding
return a.val == b.val;
}
// get all the nodes at next level, adding the spcial "padding" node when a child node is null.
ArrayList<Tree> getNextLevel(ArrayList<Tree> list, Tree padding) {
ArrayList<Tree> result = new ArrayList<>();
for (Tree node : list) {
if (node == padding) {
continue;
}
if (node.left != null) {
result.add(node.left);
} else {
result.add(padding);
}
if (node.right != null) {
result.add(node.right);
} else {
result.add(padding);
}
}
return result;
}
// this is how most people implemented. It is a better solution:
// shorter code, less memory and better performance
// performance: 85%
boolean versionB(Tree root) {
if (root == null) {
return true;
}
return recursionB(root.left, root.right);
}
boolean recursionB(Tree a, Tree b) {
if ((a == null) && (b == null)) {
return true;
}
if ((a == null) || (b == null)) {
return false;
}
if (a.val != b.val) {
return false;
}
return recursionB(a.left, b.right) && recursionB(a.right, b.left);
}
}
```

` ````
Solution in Python :
# class Tree:
# def __init__(self, val, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def solve(self, root):
def symmetry(l, r):
return (l is None and r is None) or bool(
l
and r
and l.val == r.val
and symmetry(l.left, r.right)
and symmetry(l.right, r.left)
)
return not root or symmetry(root.left, root.right)
```

## View More Similar Problems

## Queue using Two Stacks

A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que

View Solution →## Castle on the Grid

You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):

View Solution →## Down to Zero II

You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.

View Solution →## Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

View Solution →## Queries with Fixed Length

Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon

View Solution →## QHEAP1

This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element

View Solution →