101. Symmetric Tree

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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1

   / \

  2   2

/ \ / \

3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1

   / \

  2   2

   \   \

   3    3

Note:

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

判断一棵树是否是对称树

递归、迭代都写出来了，基本和答案一致，可以说是很屌了

递归比较好想

    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        return isSymmetric(root.left, root.right);
    }
    
    private boolean isSymmetric(TreeNode t1, TreeNode t2) {
        if (t1 == null && t2 == null) return true;
        if (t1 == null || t2 == null) return false;
        return t1.val == t2.val && isSymmetric(t1.left, t2.right) && isSymmetric(t1.right, t2.left);
    }

迭代 把需要对比的节点放在一起

public boolean isSymmetric(TreeNode root) {
    Queue<TreeNode> q = new LinkedList<>();
    q.add(root);
    q.add(root);
    while (!q.isEmpty()) {
        TreeNode t1 = q.poll();
        TreeNode t2 = q.poll();
        if (t1 == null && t2 == null) continue;
        if (t1 == null || t2 == null) return false;
        if (t1.val != t2.val) return false;
        q.add(t1.left);
        q.add(t2.right);
        q.add(t1.right);
        q.add(t2.left);
    }
    return true;
}