Dados dos árboles binarios, devuelve verdadero si dos árboles son espejo uno del otro, de lo contrario, devuelve falso.
Árbol espejo:
C++
// C++ code to check two binary trees are
// mirror.
#include<bits/stdc++.h>
using namespace std;
struct Node
{
int data;
Node* left, *right;
};
// inorder traversal of Binary Tree
void inorder(Node *n, vector<int> &v)
{
if (n->left != NULL)
inorder(n->left, v);
v.push_back(n->data);
if (n->right != NULL)
inorder(n->right, v);
}
// Checking if binary tree is mirror
// of each other or not.
bool areMirror(Node* a, Node* b)
{
if (a == NULL && b == NULL)
return true;
if (a == NULL || b== NULL)
return false;
// Storing inorder traversals of both
// the trees.
vector<int> v1, v2;
inorder(a, v1);
inorder(b, v2);
if (v1.size() != v2.size())
return false;
// Comparing the two arrays, if they
// are reverse then return 1, else 0
for (int i=0, j=v2.size()-1; j >= 0;
i++, j--)
if (v1[i] != v2[j])
return false;
return true;
}
// Helper function to allocate a new node
Node* newNode(int data)
{
Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return(node);
}
// Driver code
int main()
{
Node *a = newNode(1);
Node *b = newNode(1);
a -> left = newNode(2);
a -> right = newNode(3);
a -> left -> left = newNode(4);
a -> left -> right = newNode(5);
b -> left = newNode(3);
b -> right = newNode(2);
b -> right -> left = newNode(5);
b -> right -> right = newNode(4);
areMirror(a, b)? cout << "Yes" : cout << "No";
return 0;
}
Python3
# Python3 code to check two binary trees are
# mirror.
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# inorder traversal of Binary Tree
def inorder(n, v):
if (n.left != None):
inorder(n.left, v);
v.append(n.data);
if (n.right != None):
inorder(n.right, v);
# Checking if binary tree is mirror
# of each other or not.
def areMirror(a, b):
if (a == None and b == None):
return True;
if (a == None or b== None):
return False;
# Storing inorder traversals of both
# the trees.
v1 = []
v2 = []
inorder(a, v1);
inorder(b, v2);
if (len(v1) != len(v2)):
return False;
# Comparing the two arrays, if they
# are reverse then return 1, else 0
i = 0
j = len(v2) - 1
while j >= 0:
if (v1[i] != v2[j]):
return False
i+=1
j-=1
return True;
# Helper function to allocate a new node
def newNode(data):
node = Node(data)
return node
# Driver code
if __name__=="__main__":
a = newNode(1);
b = newNode(1);
a.left = newNode(2);
a.right = newNode(3);
a.left.left = newNode(4);
a.left.right = newNode(5);
b.left = newNode(3);
b.right = newNode(2);
b.right.left = newNode(5);
b.right.right = newNode(4);
if areMirror(a, b):
print("Yes")
else:
print("No")
# This code is contributed by rutvik_56
C#
// C# code to check two binary trees are
// mirror.
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
class Node
{
public int data;
public Node left, right;
};
// inorder traversal of Binary Tree
static void inorder(Node n, ref List<int> v)
{
if (n.left != null)
inorder(n.left, ref v);
v.Add(n.data);
if (n.right != null)
inorder(n.right, ref v);
}
// Checking if binary tree is mirror
// of each other or not.
static bool areMirror(Node a, Node b)
{
if (a == null && b == null)
return true;
if (a == null || b == null)
return false;
// Storing inorder traversals of both
// the trees.
List<int> v1 = new List<int>();
List<int> v2 = new List<int>();
inorder(a, ref v1);
inorder(b, ref v2);
if (v1.Count != v2.Count)
return false;
// Comparing the two arrays, if they
// are reverse then return 1, else 0
for(int i = 0, j = v2.Count - 1; j >= 0;
i++, j--)
if (v1[i] != v2[j])
return false;
return true;
}
// Helper function to allocate a new node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return(node);
}
// Driver code
static void Main(string []args)
{
Node a = newNode(1);
Node b = newNode(1);
a.left = newNode(2);
a.right = newNode(3);
a.left.left = newNode(4);
a.left.right = newNode(5);
b.left = newNode(3);
b.right = newNode(2);
b.right.left = newNode(5);
b.right.right = newNode(4);
if (areMirror(a, b))
{
Console.Write("Yes");
}
else
{
Console.Write("No");
}
}
}
// This code is contributed by pratham76
Javascript
<script>
// JavaScript code to check two binary trees are
// mirror.
class Node
{
constructor()
{
this.data = 0;
this.left = null;
this.right = null;
}
};
// inorder traversal of Binary Tree
function inorder(n, v)
{
if (n.left != null)
inorder(n.left, v);
v.push(n.data);
if (n.right != null)
inorder(n.right, v);
}
// Checking if binary tree is mirror
// of each other or not.
function areMirror(a, b)
{
if (a == null && b == null)
return true;
if (a == null || b == null)
return false;
// Storing inorder traversals of both
// the trees.
var v1 = [];
var v2 = [];
inorder(a, v1);
inorder(b, v2);
if (v1.length != v2.length)
return false;
// Comparing the two arrays, if they
// are reverse then return 1, else 0
for(var i = 0, j = v2.length - 1; j >= 0;
i++, j--)
if (v1[i] != v2[j])
return false;
return true;
}
// Helper function to allocate a new node
function newNode(data)
{
var node = new Node();
node.data = data;
node.left = node.right = null;
return(node);
}
// Driver code
var a = newNode(1);
var b = newNode(1);
a.left = newNode(2);
a.right = newNode(3);
a.left.left = newNode(4);
a.left.right = newNode(5);
b.left = newNode(3);
b.right = newNode(2);
b.right.left = newNode(5);
b.right.right = newNode(4);
if (areMirror(a, b))
{
document.write("Yes");
}
else
{
document.write("No");
}
</script>