LeetCode-in-All

101. Symmetric Tree

Easy

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

Example 1:

Input: root = [1,2,2,3,4,4,3]

Output: true

Example 2:

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

Follow up: Could you solve it both recursively and iteratively?

Solution

; Definition for a binary tree node.
#|

; val : integer?
; left : (or/c tree-node? #f)
; right : (or/c tree-node? #f)
(struct tree-node
  (val left right) #:mutable #:transparent)

; constructor
(define (make-tree-node [val 0])
  (tree-node val #f #f))

|#

(define/contract (is-symmetric root)
  (-> (or/c tree-node? #f) boolean?)
   (define (two-symmetric? left right)
    (cond
        [(and (not left) (not right )) #t]
        [(and left (not right )) #f]
        [(and (not left) right) #f]
        [(not (equal? (tree-node-val left) (tree-node-val right))) #f]
        [else 
            (and (two-symmetric? (tree-node-left left) (tree-node-right right))
                 (two-symmetric? (tree-node-right left) (tree-node-left right)))]))
  (two-symmetric? (tree-node-left root) (tree-node-right root)))

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