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.
class Solution {
public boolean isSymmetric(TreeNode root) {
if(root ==null) return true;
return isSy(root.left,root.right);
}
public boolean isSy(TreeNode s,TreeNode t) {
if(s == null || t == null) return s==t; if(s.val != t.val) return false; return isSy(s.left,t.right) && isSy(s.right,t.left);
}
}