public static bool isSymmetricTree(Node root)
{
if (root == null)
return true;
return isSymmetricTreeCore(root, root);
}
static void Main(string[] args)
{
Node root = new Node(6);
root.left = new Node(8);
root.right = new Node(8);
root.left.left = new Node(9);
root.left.right = new Node(10);
root.right.left = new Node(10);
root.right.right = new Node(7);
bool result = isSymmetricTree(root);
}
/*
* 1. empty node tree is symmetric
* 2. Base case: if the binary tree has 3 nodes, left child value should be equal to right child value
* 3. If the tree order level is more than 2, the left child of left child should be equal to right child of right child;
* the right child of left child should be equal to left child of right child
* 4. Recursively do the same thing
* 5. Check the code redundancy, and make sure that no redudant code; check the code logic
*/
public static bool isSymmetricTreeCore(Node lNode, Node rNode)
{
if (lNode == null && rNode == null)
return true; // empty tree is symmetric
if (lNode == null || rNode == null)
return false;
// base case, 3 nodes in the tree
if (lNode.data != rNode.data)
return false;
return isSymmetricTreeCore(lNode.left, rNode.right) &&
isSymmetricTreeCore(lNode.right, rNode.left);
}
