LeetCode #226 Invert Binary Tree

Posted on | Post modified | In , , , , , , , , , ,

Problem

Invert a binary tree.

Example

Input:

1
2
3
4
5
     4
   /   \
  2     7
 / \   / \
1   3 6   9

Output:

1
2
3
4
5
     4
   /   \
  7     2
 / \   / \
9   6 3   1

Solution1

Method: Recursive

Time Complexity:

Space Complexity: 

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.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if not root:
            return root

        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        self.invertTree(root.right)
        
        return root

Solution2

Method: Depth First Search

Time Complexity: O(n)

Space Complexity: O(n)

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.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if not root:
            return root

        stack = [root]
        while stack:
            node = stack.pop()
            node.left, node.right = node.right, node.left
            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)
        
        return root

or

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, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        stack = [root]
        while stack:
            node = stack.pop()
            if node:
                node.left, node.right = node.right, node.left
                stack.append(node.left)
                stack.append(node.right)
        
        return root

Solution3

Method: Breadth First Search

Time Complexity: O(n)

Space Complexity: O(n)

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, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        queue = collections.deque([root])
        while queue:
            node = queue.popleft()
            if node:
                node.left, node.right = node.right, node.left
                queue.append(node.left)
                queue.append(node.right)
        
        return root