Title Description
Downward from the print out of each node of the binary tree, with the layer node from left to right printing.
A first solution, a secondary queue implementation construct
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# 返回从上到下每个节点值列表,例:[1,2,3]
def PrintFromTopToBottom(self, root):
# write code here
if not root:
return []
queue = []
result = []
queue.append(root)
while len(queue) > 0:
node = queue.pop(0)
result.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return result
The second solution, breadth-first search
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# 返回从上到下每个节点值列表,例:[1,2,3]
def PrintFromTopToBottom(self, root):
# write code here
def helper(node, level):
if not node:
return
else:
sol[level-1].append(node.val)
if len(sol) == level: # 遍历到新层时,只有最左边的结点使得等式成立
sol.append([])
helper(node.left, level+1)
helper(node.right, level+1)
sol = [[]]
helper(root, 1)
a = []
while sol:
a += sol.pop(0)
return a