Leetcode 545. Boundary of Binary Tree (Python)

Related Topic

Depth-First-Search. Tree. Preorder-Traversal. Inorder-Traversal.

Description

Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes. (The values of the nodes may still be duplicates.)

Left boundary is defined as the path from root to the left-most node. Right boundary is defined as the path from root to the right-most node. If the root doesn’t have left subtree or right subtree, then the root itself is left boundary or right boundary. Note this definition only applies to the input binary tree, and not applies to any subtrees.

The left-most node is defined as a leaf node you could reach when you always firstly travel to the left subtree if exists. If not, travel to the right subtree. Repeat until you reach a leaf node.

The right-most node is also defined by the same way with left and right exchanged.

Sample I/O

Example 1

Input:
  1
   \
    2
   / \
  3   4

Ouput:
[1, 3, 4, 2]

Explanation:
The root doesn't have left subtree, so the root itself is left boundary.
The leaves are node 3 and 4.
The right boundary are node 1,2,4. Note the anti-clockwise direction means you should output reversed right boundary.
So order them in anti-clockwise without duplicates and we have [1,3,4,2].

Methodology

This question can be solved by Depth First Search.

This question applied various of knowledge of tree. To solve this question, we first use preorder to get the left boundary node value, then we use dfs to find the leaves node value and append to the left boundary node value. At last, we use postorder to find the right boundary and append to the previous node values.

Code

"""
# Definition for a Node.
class Node:
    def __init__(self, val, left, right, next):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
def boundaryOfBinaryTree(self, root: TreeNode) -> List[int]:
        if root is None: return []
        res=[root.val]
        def leftBoundary(root):
            if root is None or root.left is None and root.right is None: return
            res.append(root.val)
            if root.left:
                leftBoundary(root.left)
            else:
                leftBoundary(root.right)
            
        def rightBoundary(root):
            if root is None or root.left is None and root.right is None: return
            if root.right:
                rightBoundary(root.right)
            else:
                rightBoundary(root.left)
            res.append(root.val)
            
        def leaves(node):
            if node is None: return
            if node.left is None and node.right is None and node != root:
                res.append(node.val)
            leaves(node.left)
            leaves(node.right)
        
        leftBoundary(root.left)
        leaves(root)
        rightBoundary(root.right)
        return res

BigO

We traversal left subtree which takes O(n/2), right subtree which is O(n/2) and taversal all nodes for leaves and it takes O(n). In total is O(2n)