[Cracking the Coding Interview] 4.3 List of Depths

Given a binary tree, design an algorithm which creates a linked list of all the nodes at each depth.(e.g., if you have a tree with depth D, you'll have D linked lists)

这道题给定一个二叉树，要求我们对于每一层的节点建立链表。这里涉及几个数的概念，depth, height, level. 你能分得清楚吗？看以下的定义。

Height of node – The height of a node is the number of edges on the longest downward path between that node and a leaf.

• 从node到leaf会有多条路径，最长的路径的边的条数是这个node的height。

Height of tree - The height of a tree is the number of edges on the longest downward path between the root and a leaf.

• 所以有 the height of a tree is the height of its root.
• 我们经常会问这样的问题：what is the max number of nodes a tree can have if the height of the tree is h？Answer: 2^h - 1

Depth –The depth of a node is the number of edges from the node to the tree's root node.

• Depth 和 Height比是方向相反的，Height是以leaf为基准，而Depth以root为基准
• depth of root is 0.

Level – The level of a node is defined by 1 + the number of connections between the node and the root.

• level = depth + 1
• 要注意level从1开始，depth从0开始！！

这道题要求我们对tree的每一个depth的nodes建立个链表，实际上就是对数进行层序遍历，BFS是经典的层序遍历的方法，见如下代码：

class TreeNode
  attr_accessor :val, :left, :right
 
  def initialize(val)
    @val = val
    @left = nil
    @right = nil
  end
end
 
class LinkedListNode
  attr_accessor :val, :next
 
  def initialize(val)
    @val = val
    @next = nil
  end
end
 
def build_linked_lists(root)
  return [] if root.nil?
 
  lists = []
  q = Queue.new
  q << root
 
  while !q.empty?
    size = q.size
 
    dummy = LinkedListNode.new(0)
    head = dummy
 
    size.times do
      front = q.pop
      head.next = LinkedListNode.new(front.val)
      head = head.next
 
      q << front.left if front.left
      q << front.right if front.right
    end
    lists << dummy.next
  end
  lists
end