对称二叉树 没什么好说的。使用递归和迭代的思路都可
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
return isMirror(root.left, root.right);
}
public boolean isMirror(TreeNode node1, TreeNode node2){
if (node1 == null && node2 == null) return true;
if (node1 == null || node2 == null) return false;
if (node1.val != node2.val) return false;
return isMirror(node1.left, node2.right) && isMirror(node1.right, node2.left);
}
