LC-0226 Easy LeetCode

226. Invert Binary Tree

Tree Depth-First Search Breadth-First Search Binary Tree

Read the full problem statement on LeetCode.

Difficulty: easy Acceptance: 79% Topics: Tree, Depth-First Search, Breadth-First Search, Binary Tree

Reading material

Reference solution (spoiler · python)

# Time:  O(n)
# Space: O(h)

import collections


# BFS solution.
class Queue(object):
    def __init__(self):
        self.data = collections.deque()

    def push(self, x):
        self.data.append(x)

    def peek(self):
        return self.data[0]

    def pop(self):
        return self.data.popleft()

    def size(self):
        return len(self.data)

    def empty(self):
        return len(self.data) == 0

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    # @param {TreeNode} root
    # @return {TreeNode}
    def invertTree(self, root):
        if root is not None:
            nodes = Queue()
            nodes.push(root)
            while not nodes.empty():
                node = nodes.pop()
                node.left, node.right = node.right, node.left
                if node.left is not None:
                    nodes.push(node.left)
                if node.right is not None:
                    nodes.push(node.right)

        return root

# Time:  O(n)
# Space: O(h)
# Stack solution.
class Solution2(object):
    # @param {TreeNode} root
    # @return {TreeNode}
    def invertTree(self, root):
        if root is not None:
            nodes = []
            nodes.append(root)
            while nodes:
                node = nodes.pop()
                node.left, node.right = node.right, node.left
                if node.left is not None:
                    nodes.append(node.left)
                if node.right is not None:
                    nodes.append(node.right)

        return root

# Time:  O(n)
# Space: O(h)
# DFS, Recursive solution.
class Solution3(object):
    # @param {TreeNode} root
    # @return {TreeNode}
    def invertTree(self, root):
        if root is not None:
            root.left, root.right = self.invertTree(root.right), \
                                    self.invertTree(root.left)

        return root

Solution from kamyu104/LeetCode-Solutions · MIT