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.
public class Solution {
public boolean isSymmetric(TreeNode root) {
return root != null? Symmetric(root.left,root.right):true;
}
private boolean Symmetric(TreeNode L,TreeNode R)
{
if(L==null&&R==null)
return true;
if(L==null||R==null) // 已经将都是null相等的情况排除了
return false;
return L.val == R.val&&Symmetric(L.left,R.right)&&Symmetric(L.right,R.left);
}
}
