1 """ 2 Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. 3 Note: A leaf is a node with no children. 4 Example: 5 Given the below binary tree and sum = 22, 6 5 7 / 8 4 8 9 / / 10 11 13 4 11 / / 12 7 2 5 1 13 Return: 14 [ 15 [5,4,11,2], 16 [5,8,4,5] 17 ] 18 """ 19 """ 20 此题好题：写三种解法 21 解法一：recursive 22 """ 23 class TreeNode: 24 def __init__(self, x): 25 self.val = x 26 self.left = None 27 self.right = None 28 class Solution1: 29 def pathSum(self, root, sum): 30 if not root: 31 return [] 32 res = [] 33 path = [] 34 self.recursive(root, res, sum, path) 35 return res 36 def recursive(self, root, res, sum, path): 37 if not root.left and not root.right and sum == root.val: 38 path.append(root.val) 39 res.append(path) 40 if root.left: 41 self.recursive(root.left, res, sum-root.val, path+[root.val]) 42 if root.right: 43 self.recursive(root.right, res, sum-root.val, path+[root.val]) 44 """ 45 解法二：DFS 46 """ 47 class Solution2: 48 def pathSum(self, root, s): #注意这里,将原本的sum改为s了 49 if not root: 50 return [] 51 stack = [] 52 res = [] 53 stack.append((root, [root.val])) 54 while stack: 55 root, path = stack.pop() 56 if not root.left and not root.right and sum(path) == s: #！！！关键点 57 res.append(path) 58 if root.left: 59 stack.append((root.left, path+[root.left.val])) 60 if root.right: 61 stack.append((root.right, path+[root.right.val])) 62 return res 63 """ 64 解法三：BFS 65 """ 66 class Solution3: 67 def pathSum(self, root, s): 68 if not root: 69 return [] 70 queue = [] 71 res = [] 72 queue.append((root, [root.val])) 73 while queue: 74 root, path = queue.pop(0) #队列和栈的区别在这 75 if not root.left and not root.right and sum(path) == s: 76 res.append(path) 77 if root.left: 78 queue.append((root.left, path+[root.left.val])) 79 if root.right: 80 queue.append((root.right, path+[root.right.val])) 81 return res 82 if __name__ == '__main__': 83 l1 = TreeNode(5) 84 l2 = TreeNode(4) 85 l3 = TreeNode(8) 86 l4 = TreeNode(11) 87 l5 = TreeNode(5) 88 l6 = TreeNode(13) 89 l7 = TreeNode(4) 90 l8 = TreeNode(7) 91 l9 = TreeNode(2) 92 l10 = TreeNode(1) 93 l1.left = l2 94 l1.right = l3 95 l2.left = l4 96 l3.left = l6 97 l3.right = l7 98 l7.left = l5 99 l4.left = l8 100 l4.right = l9 101 l7.right = l10 102 ans = Solution3() 103 print(ans.pathSum(l1, 22))