CareerCup

Do level order traversal. At each level, check if the elements form a palindrome.

n must be even.

Use a subroutine , isMirrorImage( Tree a, Tree b )

isSymmetric ( Tree r )
{
	children[]  child = r.children;
	
        for ( s = 0; e = n -1;   )
        {
		isMirrorImage ( child [ s ], child [ e ] );
		s ++; e --;
	}


}

#include <iostream>
#include <vector>
using namespace std;

struct TreeNode
{
	int val;
	vector<TreeNode*> children;
	TreeNode(int val)
	{
		this->val = val;
	}
};

bool IsMirrorImage(TreeNode* t1, TreeNode* t2)
{
	if(t1 == NULL && t2 == NULL)
	{
		return true;
	}
	if(t1 == NULL || t2 == NULL)
	{
		return false;
	}
	bool isMirrorImage = t1->val == t2->val && (t1->children).size() == (t2->children).size();
	int size = (t1->children).size();
	for(int i=0; isMirrorImage == true && i < size; i++)
	{
		isMirrorImage = IsMirrorImage((t1->children)[i], (t2->children)[size-i-1]);
	}
	return isMirrorImage;
}
int main() {
	TreeNode* root1 = new TreeNode(1);
	root1->children.push_back(new TreeNode(2));
	root1->children.push_back(new TreeNode(3));
	TreeNode* root2 = new TreeNode(1);
	root2->children.push_back(new TreeNode(3));
	root1->children.push_back(new TreeNode(2));
	cout<<IsMirrorImage(root1, root2);
}