• 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:

     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

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  • 原文地址:https://www.cnblogs.com/NPC-assange/p/9622423.html
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