劍指Offer24:二叉樹中和為某一值的路徑
阿新 • • 發佈:2018-12-11
思路:
實在沒能搞懂大佬們的思路,坐等第二遍刷再看吧
# -*- coding:utf-8 -*- # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: # 返回二維列表,內部每個列表表示找到的路徑 def FindPath(self, root, expectNumber): if not root: return [] tmp = [] if not root.left and not root.right and root.val == expectNumber: return [[root.val]] else: left = self.FindPath(root.left,expectNumber-root.val) right = self.FindPath(root.right,expectNumber-root.val) for item in left+right: tmp.append([root.val]+item) return tmp