【剑指offer】Q25：二叉树中和为某一值的路径

来源：互联网发布：如何做读书笔记知乎编辑：程序博客网时间：2024/06/05 00:19

说明：最烦的就是看别人的博客，题解里直接上代码，一行分析都没有，不过这个题。。。快哭了

class BTNode():def __init__(self, val = -1):self.val = valself.left = Noneself.right = Noneclass BTree():def __init__(self):self.root = None'''ex               1              / \             2   3            /   /           4   5 treeArray = [1,2,3,4,'#',5]'''def createTree(self, treeArray):self._createTree(treeArray)def _createTree(self,treeArray, i = 0):if i > len(treeArray) :return Noneif treeArray[i] == '#':return Noneroot = BTNode(int(treeArray[i]))if self.root == None:self.root = root#create left branchl = 2*i + 1if l < len(treeArray):root.left = self._createTree(treeArray, l)#create right branchr = 2*i + 2if r < len(treeArray):root.right = self._createTree(treeArray,r)return root    def preorder(self, root):if root == None:returnprint root.valself.preorder(root.left)self.preorder(root.right)def inorder(self, root):if root == None:returnself.inorder(root.left)print root.valself.inorder(root.right)def postorder(self, root):if root == None:returnself.postorder(root.left)self.postorder(root.right)print root.val

def Print(path):for i in range(len(path)):print path[i]print "--------------------"def pathSum(broot, remainder, path):if broot == None:returnif remainder < broot.val:returnpath.append(broot.val)remainder -= broot.valif remainder == 0:if broot.left == None and broot.right == None:Print(path)else:returnpathSum(broot.left, remainder, path)pathSum(broot.right, remainder,path)path = path.pop()if __name__ == '__main__':array = [10,5,12,4,7]bt = BTree()bt.createTree(array)#bt.preorder(bt.root)path = []remainder = 22pathSum(bt.root, remainder, path)

0 0