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.
Bonus points if you could solve it both recursively and iteratively.
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class Solution { | |
public boolean isSymmetric(TreeNode root) { | |
// corner cases | |
if (root == null) { | |
return true; | |
} | |
return isSymmetric(root.left, root.right); | |
} | |
public boolean isSymmetric(TreeNode left, TreeNode right) { | |
// corner cases | |
if (left == null && right == null) { | |
return true; | |
} | |
if (left == null && right != null) { | |
return false; | |
} | |
if (left != null && right == null) { | |
return false; | |
} | |
return left.val == right.val && isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left); | |
} | |
} |
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