freewheel笔试的一题（二叉树最小耗时）

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param root TreeNode类 
# @param failId int整型 
# @param timeCost int整型一维数组 
# @return int整型
#
class Solution:
    def GetMinTimeCost(self , root , failId , timeCost ):
        # write code here
        if not root:
            return 0
        #首先根据编码找到fail节点：节点标号从0开始，节点编码唯一且连续
        def preorder(root,failid):
            if not root:
                return None
            if root.val==failid:
                return root
            else:
                preorder(root.left,failid)
                preorder(root.right,failid)
        a=preorder(root, failId)
        if not a:
            return 0
        def cost(root,timeCost,res):
            #应该是每一步都要求是最小的
            queue=[root]
            while queue:
                n=len(queue)
                l=[]
                for i in range(n):
                    tmp=queue.pop(0)
                    if tmp.left==None and tmp.right==None:
                        continue
                    if tmp.left  or tmp.right:
                        l.append(timeCost[tmp.val])
                        if tmp.right:
                            queue.append(tmp.right)
                        if tmp.left:
                            queue.append(tmp.left)
                    
                if not l:
                    res+=0
                else:
                    res+=min(l)
            return res
        h=cost(a,timeCost,0)
        return h