# 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);
}

// 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

ArrayList<Tree> list = new ArrayList<>();

while (!list.isEmpty()) {
return false;
}
}
return true;
}

// check as if we were checking palindrome: all the nodes at one level
boolean isValid(ArrayList<Tree> list, Tree padding) {
int tail = list.size() - 1;

return false;
}
tail--;
}

return true;
}

boolean isValid(Tree a, Tree b, Tree padding) {
return true;
}
return false;
}
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) {
continue;
}
if (node.left != null) {
} else {
}

if (node.right != null) {
} else {
}
}
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)```
```

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