Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:
1
/ \
2 2
/ \ / \
3 4 4 3
Input: root = [1,2,2,3,4,4,3]
Output: true
Example 2:
1
/ \
2 2
\ \
3 3
Input: root = [1,2,2,null,3,null,3]
Output: false
Constraints:
- The number of nodes in the tree is in the range [1, 1000].
- -100 <= Node.val <= 100
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
|
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isSymmetric(self, root: Optional[TreeNode]) -> bool:
def checkSymmetry(
left: Optional[TreeNode],
right: Optional[TreeNode]
) -> bool:
if not left or not right:
return left == right
if left.val != right.val:
return False
return (
checkSymmetry(left.left, right.right)
and checkSymmetry(left.right, right.left)
)
return checkSymmetry(root.left, root.right)
|
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.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func isSymmetric(root *TreeNode) bool {
return root != nil && checkSymmetry(root.Left, root.Right)
}
func checkSymmetry(left *TreeNode, right *TreeNode) bool {
if left == nil || right == nil {
return left == right
}
if left.Val != right.Val {
return false
}
return checkSymmetry(left.Left, right.Right) && checkSymmetry(left.Right, right.Left)
}
|
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
|
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
return (root == null) || help(root.left, root.right);
}
private boolean help(TreeNode l, TreeNode r){
if (l == null || r == null) return l == r;
if(l.val != r.val) return false;
return help(l.left, r. right) && help(l.right, r.left);
}
}
|