面试题34：二叉树的和为某一值的路径

Posted 2022-07-15 ivyharding

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
import copy
class Solution:
    # 返回二维列表，内部每个列表表示找到的路径
    def FindPath(self, root, expectNumber):
        # write code here
        # 首先判断边界
        if root == None:
            return []
        
        # 采用广度优先遍历我们的树，广度优先中有构造一个辅助列表
        support = [root]
        supportArray = [[root.val]]
        retArray = []
        
        while support:
            temp = support[0]
            tempsupportArray = supportArray[0]
            # 需要判断当前节点是否为叶子节点
            if temp.left==None and temp.right==None:
                tempSum = sum(tempsupportArray)
                if tempSum==expectNumber:
                    retArray.insert(0,tempsupportArray)
            if temp.left:
                support.append(temp.left)
                #这里为啥用copy????
                newtempsupportArray = copy.copy(tempsupportArray)
                newtempsupportArray.append(temp.left.val)
                supportArray.append(newtempsupportArray)
            if temp.right:
                support.append(temp.right)
                newtempsupportArray = copy.copy(tempsupportArray)
                newtempsupportArray.append(temp.right.val)
                supportArray.append(newtempsupportArray)
                
            del supportArray[0]
            del support[0]
        return retArray