101-symmetric-tree

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Description

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(Python)

# 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:
        return self.iterative(root)

    # Time Complexity: O(n)
    # Space Complexity: O(h)
    def recursive(self, root1, root2=None):
        if root1 is None and root2 is None:
            return True

        if root1 is not None and root2 is not None:
            if root1.val == root2.val:
                return self.recursive(root1.left, root2.right) and self.recursive(
                    root1.right, root2.left
                )

        return False

    # Time Complexity: O(n)
    # Space Complexity: O(n)
    def iterative(self, root):
        q = deque([])
        q.append(root)
        q.append(root)

        while q:
            original_node = q.popleft()
            mirrored_node = q.popleft()

            if original_node.val != mirrored_node.val:
                return False

            if original_node.left and mirrored_node.right:
                q.append(original_node.left)
                q.append(mirrored_node.right)
            elif original_node.left or mirrored_node.right:
                return False

            if original_node.right and mirrored_node.left:
                q.append(original_node.right)
                q.append(mirrored_node.left)
            elif original_node.right or mirrored_node.left:
                return False

        return True