Python3解leetcode Path Sum III

问题描述：

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

思路：

仍然是深度优先遍历的原则，该方法需要多做题多巩固啊！！占了我一整天时间的一道题。

从头到尾遍历每一个节点，记录从根节点到每一个节点的sum。

通过查找当前节点和目标sum之间的差值，进行相应剪枝动作

当退出当前层次时候，需要减去当前层次计算出来的和，故每次return之前都有一个减一的动作

代码：

 # Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None

 class Solution:
     def pathSum(self, root: TreeNode, sum: int) -> int:
         if root == None : return 0
         prefix = {0:1}
         return self.Calc(root,0,sum,prefix)

     def Calc(self,root,cursum,target,prefix):
         if root == None : return 0
         cursum += root.val
         res = prefix.get(cursum - target,0)
         prefix[cursum] = prefix.get(cursum,0) + 1
         res += self.Calc(root.left,cursum,target,prefix)  + self.Calc(root.right,cursum,target,prefix)
         prefix[cursum] = prefix.get(cursum) - 1
         return res