输入一棵二叉树和一个整数，打印出二叉树中节点值的和为输入整数的所有路径。从树的根节点开始往下一直到叶节点所经过的节点形成一条路径。
示例: 给定如下二叉树，以及目标和 sum = 22，
5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1返回:
[ [5,4,11,2], [5,8,4,5] ] # encoding=utf-8 # https://www.cnblogs.com/dongyangblog/p/11210820.html # 0 # 1 2 # 3 4 5 6 class BinaryTree: def __init__(self, data): self.data = data self.left = None self.right = None def get(self): return self.data def getleft(self): return self.left def getright(self): return self.right def setleft(self, node): self.left = node def setright(self, node): self.right = node binaryTree = BinaryTree(0) binaryTree.setleft(BinaryTree(1)) binaryTree.setright(BinaryTree(2)) binaryTree.getleft().setleft(BinaryTree(3)) binaryTree.getleft().setright(BinaryTree(4)) binaryTree.getright().setleft(BinaryTree(5)) binaryTree.getright().setright(BinaryTree(6)) def pathSum(root, sum): res, path = [], [] help(root, sum, res, path) return res def help(root, sum, res, path): if not root: return path.append(root.data) sum -= root.data if sum == 0 and not root.left and not root.right: res.append(list(path)) help(root.left, sum, res, path) help(root.right, sum, res, path) path.pop() print(pathSum(binaryTree, 7)) def pathSum(root, sum): res, path = [], [] help(root, sum, res, path) return res def help(root, sum, res, path): if not root: return path.append(root.data) sum -= root.data if sum == 0 and not root.left and not root.right: res.append(list(path)) help(root.left, sum, res, path) help(root.right, sum, res, path) path.pop() print(pathSum(binaryTree, 7))[[0, 2, 5]]
