Question

Code using JAVA:

must include a "Main Method" to run on "intelliJ".

Hint: Use recursion

"

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isSymmetric(TreeNode root) {
  
}
}

"

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

Follow up: Solve it both recursively and iteratively.

Answer 1

class Node

{

int k;

Node l, r;

Node(int item)

{

k= item;

l = r = null;

}

}

class BinaryTree

{

Node root;

// returns true if trees with roots as root1 and root2 are mirror

boolean isMirror(Node node1, Node node2)

{

// if both trees are empty, then they are mirror image

if (node1 == null && node2 == null)

return true;

  

if (node1 != null && node2 != null && node1.k == node2.k)

return (isMirror(node1.l, node2.r)

&& isMirror(node1.r, node2.l));

  

return false;

}

  

boolean isSymmetric(Node node)

{

// check if tree is mirror of itself

return isMirror(root, root);

}

// Driver program

public static void main(String args[])

{

BinaryTree tree = new BinaryTree();

tree.root = new Node(1);

tree.root.l = new Node(2);

tree.root.r = new Node(2);

tree.root.l.l = new Node(3);

tree.root.l.r = new Node(4);

tree.root.r.l = new Node(4);

tree.root.r.r = new Node(3);

boolean output = tree.isSymmetric(tree.root);

if (output == true)

System.out.println("1");

else

System.out.println("0");

}

  

}