Python编程求解二叉树中和为某一值的路径代码示例

# -*- coding:utf-8 -*- 
class TreeNode(): 
  def __init__(self,x): 
    self.val = x 
    self.left = None 
    self.right = None 
 
def function(root,target_number): 
  result = [] 
  if not root: 
    return result 
#  如果只有根节点或者找到叶子节点，我们就把其值返回 
  if not root.left and not root.right and root.val == target_number: 
    return [[root.val]] 
  else: 
#  如果不是叶子节点，我们分别对根节点的左子树、右子树进行递归，注意修改变量: 
    left = function(root.left,target_number - root.val) 
    right = function(root.right,target_number - root.val) 
    for item in left+right： 
      result.append([root.val]+item) 
    return result