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

For example, this binary tree is symmetric:

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

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

判斷二叉樹是否是平衡樹，比如有兩個節點n1, n2，我們需要比較n1的左子節點的值和n2的右子節點的值是否相等，同時還要比較n1的右子節點的值和n2的左子結點的值是否相等，以此類推比較完所有的左右兩個節點。我們可以用遞迴和迭代兩種方法來實現，寫法不同，但是演算法核心都一樣。

public boolean isSymmetric(TreeNode root) {
    if (root == null) {
        return true;
    }
    return isSameTree(root.left, root.right);
}

public boolean isSameTree(TreeNode p, TreeNode q) {
    if (p == null || q == null) {
        return p == q;
    }
    boolean f = p.val == q.val;
    boolean f2 = isSameTree(p.left, q.right);
    boolean f3 = isSameTree(p.right, q.left);
    return f && f2 && f3;
}
 

類似問題:

 1. 222Count Complete Tree

2. 110Balanced Binary Tree