101Symmetric Tree

Solution 1: Recursive O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isMirror(TreeNode t1, TreeNode t2){
        //If Both are null;
        if(t1 == null && t2 == null){
            return true;
        }
        //If only one is null
        else if(t1 == null || t2 == null){
            return false;
        }
        //If both are not null
        else{
            return (t1.val == t2.val) && isMirror(t1.left, t2.right) && isMirror(t1.right, t2.left);
        }
    }
    public boolean isSymmetric(TreeNode root) {
        return isMirror(root, root);
    }
}

Solution 2 :Using Queue O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null){
            return true;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root.left);
        queue.add(root.right);
        while(!queue.isEmpty()){
            TreeNode left = queue.poll();
            TreeNode right = queue.poll();
            //Should Use continue instead of return true;
            if(left == null && right == null){
                continue;
            }
            else if(left == null || right == null){
                return false;
            }
            else{
                if(left.val != right.val){
                    return false;
                }
                queue.add(left.left);
                queue.add(right.right);
                queue.add(left.right);
                queue.add(right.left);
            }
        }
        return true;
    }
}