输入一颗二叉树和一个整数，打印出二叉树中节点值的和为输入整数的所有路径。路径定义为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

class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None def findPath(root, n): if not root: return [] result = [] def findPath2(root, path, currentNum): currentNum += root.val path.append(root) # 判断root是否为叶子节点 flag = root.left == None and root.right == None if currentNum == n and flag: onepath = [] for node in path: onepath.append(node.val) result.append(onepath) if currentNum < n: if root.left: findPath2(root.left, path, currentNum) if root.right: findPath2(root.right, path, currentNum) path.pop() findPath2(root, [], 0) return result root = TreeNode(1)
left = TreeNode(2)
right = TreeNode(3)
root.left = left
root.right = right

left1 = TreeNode(4)
right1 = TreeNode(5)
right.left = left1
right.right1 = right1

left2 = TreeNode(6)
left1.left = left2

left3 = TreeNode(11)
left.left = left3

''' 1 2 3 4   5 6
'''
print(findPath(root, 14))

  
 

  
1
  
2
  
3
  
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5
  
6
  
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11
  
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54
 

[[1, 2, 11], [1, 3, 4, 6]]

  
 

  
1
 

92 - 青蛙跳台阶

文章来源: ruochen.blog.csdn.net，作者：若尘，版权归原作者所有，如需转载，请联系作者。

原文链接：ruochen.blog.csdn.net/article/details/105218428