Problem
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
Example
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.
Solution
Recursion
class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
return dfs(root.left, root.right);
}
private boolean dfs(TreeNode left, TreeNode right) {
if (left == null) return right == null;
if (right == null) return left == null;
if (left.val != right.val) {
return false;
}
return dfs(left.left, right.right) && dfs(left.right, right.left);
}
}
Iteration
class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root.left);
queue.offer(root.right);
while (!queue.isEmpty()) {
TreeNode left = queue.poll();
TreeNode right = queue.poll();
if (left == null ^ right == null) return false;
if (left == null && right == null) continue;
if (left.val != right.val) return false;
queue.offer(left.left);
queue.offer(right.right);
queue.offer(left.right);
queue.offer(right.left);
}
return true;
}
}
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