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
3
4
5
| 1
/ \
2 2
/ \ / \
3 4 4 3
|
But the following [1,2,2,null,3,null,3] is not:
O(n) time, O(n) space DFS
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
|
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
return isMirror(root.left, root.right);
}
public boolean isMirror(TreeNode left, TreeNode right) {
if (left == null || right == null)
return left == right;
return left.val == right.val && isMirror(left.left, right.right) && isMirror(left.right, right.left);
}
}
|
O(n) time O(n) space BFS
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
|
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root.left);
queue.add(root.right);
while (! queue.isEmpty()) {
TreeNode n1 = queue.poll();
TreeNode n2 = queue.poll();
if (n1 == null || n2 == null ) {
if (n1 != n2)
return false;
} else {
if (n1.val != n2.val) {
return false;
}
queue.add(n1.left);
queue.add(n2.right);
queue.add(n1.right);
queue.add(n2.left);
}
}
return true;
}
}
|