python

原题题目自行搜索

题解

思路：

dfs二叉树构建+层序遍历二叉树

• dfs深度搜优先递归每个节点的值，同时根据dfs带来的前序遍历效果==构建二叉树==
• 二叉树层序遍历

n = int(input())
leaves = list(map(int,input().split()))

class Tree():
    def __init__(self,val):
        self.left = None
        self.right = None
        self.val = val

c= []#测试使用，用于打印
#深度优先搜索的过程，顺便构造二叉树
def dfs(left,right,sumroot):
  
    if left>right:
        return 
    mid = (left+right)//2
    avg = (min(leaves[left:right+1])+max(leaves[left:right+1]))//2
    val = avg-sumroot
    c.append(val)
    node = Tree(val)
    if left==right:
        return Tree(val)

    node.left = dfs(left,mid,sumroot+val)
    node.right = dfs(mid+1,right,sumroot+val)
    return node

# root即为构造好的二叉树
root = dfs(0,len(leaves)-1,0)

#常规的二叉树层序遍历
from collections import deque
stack = deque()
stack.append(root)
res = []
while stack:
    node = stack.popleft()
    res.append(node.val)
    if node.left:
        stack.append(node.left)
    if node.right:
        stack.append(node.right)
#result
print(' '.join([str(item) for item in res]))