Leetcode - Episode 15 - Three Tree Qs (3 x E)

I felt like it was time to knock some of these tree questions out. I find that solving tree questions in Python leads to simple, elegant code.

I went over Heap's Algorithm for solving permutation problems earlier. It hasn't fully clicked with me yet though. More studying is required -- as well as checking out alternative solutions. Heap's feels a little heavy-handed to me.

Invert Binary Tree

Problem: Invert a binary tree.

I like the check I use here before I invert -- if node.left or node.right:. It's instantly obvious that it will work for all three siutations, two leafs, one leaf, None.

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

class Solution(object):
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        def recurse(node):
            if node.left or node.right:
                node.left, node.right = node.right, node.left
            if node.left:
                recurse(node.left)
            if node.right:
                recurse(node.right)
            return
        
        if root:
            recurse(root)
        return root

This problem is a little bit of a meme. I expected it to be harder than it was but I have been revising tree algorithms recently..

Runtime complexity: O(n).

Spacetime complexity: O(n).

N-ary Tree Postorder Traversal

Problem: Return the postorder traversal of a tree's values.

To think about this in another way, return the values as you would read the tree from the bottom up line by line.

"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, children):
        self.val = val
        self.children = children
"""
class Solution(object):
    def postorder(self, root):
        """
        :type root: Node
        :rtype: List[int]
        """
        vals = []
        queue = []
        if root:
            queue.append(root)
        while queue:
            cur = queue.pop()
            vals.append(cur.val)
            queue.extend(cur.children)
        
        return vals[::-1]

I recently learned (or, confirmed) that Python doesn't create new memory when slicing or creating an iterator, e.g. [1,2][::-1] / reversed([1,2]). It seems obvious now that I think about it.

Runtime complexity: O(n).

Space complexity: O(n).

N-ary Tree Preorder Traversal

Problem: Return the preorder traversal of a tree's values.

Another way to think about preorder traversal is Depth-First-Search order.

"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, children):
        self.val = val
        self.children = children
"""
class Solution(object):
    def preorder(self, root):
        """
        :type root: Node
        :rtype: List[int]
        """
        dfs_order = []
        
        def recurse(node):
            dfs_order.append(node.val)
            if node.children is not None:
                for i in range(0, len(node.children)):
                    recurse(node.children[i])
        if root:
            recurse(root)
        
        return dfs_order

I think this is a fairly terse but understandable solution but you can only make DFS so complicated.

Runtime complexity: O(n).

Spacetime complexity: O(n).