问题描述：
给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有节点值相加等于目标和。
说明: 叶子节点是指没有子节点的节点。
示例:
给定如下二叉树，以及目标和 sum = 22，
5
/
4 8
/ /
11 13 4
/
7 2 1
返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。
错误：
1 class Solution(object): 2 def hasPathSum(self, root, sum): 3 """ 4 :type root: TreeNode 5 :type sum: int 6 :rtype: bool 7 """ 8 def preOrder(self,root,sum,temp_sum): 9 if self.flag: 10 return self.flag 11 elif root and not self.flag: 12 temp_sum += root.val 13 if not root.left and not root.right: 14 if sum == temp_sum: 15 self.flag = True 16 return 17 else : 18 temp_sum -= root.val 19 self.flag = False 20 if root == None: 21 return 22 preOrder(self,root.left,sum,temp_sum) 23 preOrder(self,root.right,sum,temp_sum) 24 self.flag = False 25 return preOrder(self,root,sum,0)
改正：
1 class Solution(object): 2 def hasPathSum(self, root, sum): 3 """ 4 :type root: TreeNode 5 :type sum: int 6 :rtype: bool 7 """ 8 self.flag = False 9 def preOrder(self,root,sum,temp_sum): 10 if not self.flag: 11 if root: 12 temp_sum += root.val 13 if root.left or root.right: 14 preOrder(self,root.left,sum,temp_sum) 15 preOrder(self,root.right,sum,temp_sum) 16 else: 17 if sum == temp_sum: 18 self.flag = True 19 else: 20 temp_sum -= root.val 21 if root: 22 preOrder(self,root,sum,0) 23 return self.flag
参考：
1 class Solution(object): 2 def hasPathSum(self, root, sum): 3 """ 4 :type root: TreeNode 5 :type sum: int 6 :rtype: bool 7 """ 8 self.flag = False 9 def dfs(node,sumNow = 0,target = sum): 10 if not self.flag: 11 if node: 12 sumNow += node.val 13 if node.left or node.right: 14 dfs(node.left, sumNow, target) 15 dfs(node.right, sumNow, target) 16 else: 17 if sumNow == target: 18 self.flag = True 19 if root: 20 dfs(root) 21 return self.flag
官方：
1 class Solution(object): 2 def hasPathSum(self, root, sum): 3 """ 4 :type root: TreeNode 5 :type sum: int 6 :rtype: bool 7 """ 8 if root is None: 9 return False 10 if root.left is None and root.right is None: 11 return root.val == sum 12 if root.left == None: 13 return self.hasPathSum(root.right,sum - root.val) 14 if root.right == None: 15 return self.hasPathSum(root.left,sum - root.val) 16 return self.hasPathSum(root.left,sum - root.val) or self.hasPathSum(root.right,sum - root.val)
2018-09-10 20:54:47