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推廣網站開發,包含 Laravel 和 Kotlin 後端撰寫、自動化測試、讀書心得等。Taiwan Kotlin User Group 管理員。

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Hi, here's your problem today. This problem was recently asked by Google:

Given a binary tree, remove the nodes in which there is only 1 child, so that the binary tree is a full binary tree.

So leaf nodes with no children should be kept, and nodes with 2 children should be kept as well.

Here's a starting point:

from collections import deque

class Node(object):
  def __init__(self, value, left=None, right=None):
    self.left = left
    self.right = right
    self.value = value
  def __str__(self):
    q = deque()
    q.append(self)
    result = ''
    while len(q):
      num = len(q)
      while num > 0:
        n = q.popleft()
        result += str(n.value)
        if n.left:
          q.append(n.left)
        if n.right:
          q.append(n.right)
        num = num - 1
      if len(q):
        result += "\n"

    return result

def fullBinaryTree(node):
  # Fill this in.

# Given this tree:
#     1
#    / \ 
#   2   3
#  /   / \
# 0   9   4

# We want a tree like:
#     1
#    / \ 
#   0   3
#      / \
#     9   4

tree = Node(1)
tree.left = Node(2)
tree.right = Node(3)
tree.right.right = Node(4)
tree.right.left = Node(9)
tree.left.left = Node(0)
print fullBinaryTree(tree)
# 1
# 03
# 94