[Leetcode] 545. Boundary of Binary Tree

Update: Hmm, I feel more comfortable of doing normal problems now. Seems that keeping doing these problems will soon be limited. Once I become comfortable of doing something, I just don't feel like doing it all the time any more... Luckily, computer science is such a broad subject and it is easy to improve anywhere anytime. But then, improving on new skillsets or working on harder problems will be too hard again.. I will struggle over them for a long time.. Ahh, I guess I will just live this kind of depressed life for the rest of my life...

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.

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.

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].

Example 2

Input:
    ____1_____
   /          \
  2            3
 / \          / 
4   5        6   
   / \      / \
  7   8    9  10  
       
Ouput:
[1,2,4,7,8,9,10,6,3]

Explanation:
The left boundary are node 1,2,4. (4 is the left-most node according to definition)
The leaves are node 4,7,8,9,10.
The right boundary are node 1,3,6,10. (10 is the right-most node).
So order them in anti-clockwise without duplicate nodes we have [1,2,4,7,8,9,10,6,3].

Idea is that the boundary is made of left arm, all the leaves, and right arm. Then, just find all of them. 

Here is the code:

# 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 boundaryOfBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root: return []
        if not root.right and not root.left: return [root.val]
        
        res = [root.val]
        m = []
        self.find_leaves(root, m)
        l = []
        self.find_left(root.left, l)
        r = []
        self.find_right(root.right, r)
        #print m, l, r
        return res + l + m + r[::-1]
    
    def find_leaves(self, node, m):
        if not node:
            return 
        elif not node.right and not node.left:
            m.append(node.val)
            return
        else:
            self.find_leaves(node.left, m)
            self.find_leaves(node.right, m)
            
    def find_left(self, node, l):
        if not node:
            return
        if not node.left and not node.right:
            return
        l.append(node.val)
        if not node.left:
            self.find_left(node.right, l)
        else: self.find_left(node.left, l)
        
    def find_right(self, node, r):
        if not node:
            return
        if not node.left and not node.right:
            return
        r.append(node.val)
        if not node.right:
            self.find_right(node.left, r)
        else:
            self.find_right(node.right, r)