LeetCode--112--路径总和

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：

 class Solution(object):
     def hasPathSum(self, root, sum):
         """
         :type root: TreeNode
         :type sum: int
         :rtype: bool
         """
         def preOrder(self,root,sum,temp_sum):
             if self.flag:
                 return self.flag
             elif root and not self.flag:
                 temp_sum += root.val
                 if not root.left and not root.right:
                     if sum == temp_sum:
                         self.flag = True
                         return
                     else :
                         temp_sum -= root.val
                         self.flag = False
             if root == None:
                 return
                 preOrder(self,root.left,sum,temp_sum)
                 preOrder(self,root.right,sum,temp_sum)
         self.flag = False
         return preOrder(self,root,sum,0)

改正：

 class Solution(object):
     def hasPathSum(self, root, sum):
         """
         :type root: TreeNode
         :type sum: int
         :rtype: bool
         """
         self.flag = False
         def preOrder(self,root,sum,temp_sum):
             if not self.flag:
                 if root:
                     temp_sum += root.val
                     if root.left or root.right:
                         preOrder(self,root.left,sum,temp_sum)
                         preOrder(self,root.right,sum,temp_sum)
                     else:
                         if sum == temp_sum:
                             self.flag = True
                         else:
                             temp_sum -= root.val
         if root:
             preOrder(self,root,sum,0)
         return self.flag

参考：

 class Solution(object):
     def hasPathSum(self, root, sum):
         """
         :type root: TreeNode
         :type sum: int
         :rtype: bool
         """
         self.flag = False
         def dfs(node,sumNow = 0,target = sum):
             if not self.flag:
                 if node:
                     sumNow += node.val
                     if node.left or node.right:
                         dfs(node.left, sumNow, target)
                         dfs(node.right, sumNow, target)
                     else:
                         if sumNow == target:
                             self.flag = True
         if root:
             dfs(root)
         return self.flag

官方：

 class Solution(object):
     def hasPathSum(self, root, sum):
         """
         :type root: TreeNode
         :type sum: int
         :rtype: bool
         """
         if root is None:
             return False
         if root.left is None and root.right is None:
             return root.val == sum
         if root.left == None:
             return self.hasPathSum(root.right,sum - root.val)
         if root.right == None:
             return self.hasPathSum(root.left,sum - root.val)
         return self.hasPathSum(root.left,sum - root.val) or self.hasPathSum(root.right,sum - root.val)

2018-09-10 20:54:47