Symmetric Binary Tree

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

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

The above binary tree is symmetric.
But the following is not:

    1
   / \
  2   2
   \   \
   3    3

Return 0 / 1 ( 0 for false, 1 for true ) for this problem

Method:

DFS

Solution:

Time: O(n)
Space: O(1)

/**
 * Definition for binary tree
 * class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) {
 *      val = x;
 *      left=null;
 *      right=null;
 *     }
 * }
 */
public class Solution {
    public int isSymmetric(TreeNode A) {
        if (A == null) return 1;
        return helper(A.left, A.right);
    }
    
    private int helper(TreeNode left, TreeNode right) {
        if (left == null && right == null) return 1;
        if (left == null || right == null) return 0;
        if (left.val != right.val) return 0;
        return Math.min(helper(left.left, right.right), helper(left.right, right.left));
    }
}