打印树路径

Question

我正在学习python作为我的项目要求之一是打印二叉树。

我想出我的代码来打印树的路径。 它只停留在根节点。 我想找到我搞乱的地方。

树：

 A 
    /\ 
    B C 
//\ 
D E F 

我想输出是：

ABD

ACE

ACF

class Node(object): 
    def __init__(self, data): 
     self.data = data 
     self.children = [] 
     self.val = data 
     self.left = None 
     self.right = None 


r = Node('A') 
r.left = Node('B') 
r.right = Node('C') 
r.left.left = Node('D') 
r.right.right = Node('E') 

以上应r.right.left =节点（ 'E） r.right.right =节点（' F“）

def binaryTreePaths(root): 
     results = [] 
     c = [] 
     binary_tree_paths(root, c, results) 
     return results 

def binary_tree_paths(root, cand, res): 
    if root is None: 
     return 
    else: 
     cand.append(str(root.val)+" ") 
     if root.left is None and root.right is None: 
      p = ''.join(map(str, cand)) 
      res.append(p) 
     binary_tree_paths(root.left, cand, res) 
     binary_tree_paths(root.right, cand, res) 
     cand.pop() 

print binaryTreePaths(r) 

Answer 1

你应该有r.right.left = Node('E')，而不是r.right.right = Node('E')