leetcode 437. Path Sum III

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.

Example:

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

解法1，DFS：

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

class Solution(object):
    def pathSum(self,root,sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int
        # java solution
        public class Solution {
            public int pathSum(TreeNode root,int sum) {
                if (root == null) return 0;
                return pathSumFrom(root,sum) + pathSum(root.left,sum) + pathSum(root.right,sum);
            }

            private int pathSumFrom(TreeNode node,int sum) {
                if (node == null) return 0;
                return (node.val == sum ? 1 : 0) 
                    + pathSumFrom(node.left,sum - node.val) + pathSumFrom(node.right,sum - node.val);
            }
        }
        """
        def dfs(node,target):      
            if not node: return 0 
            return int(node.val == target)+dfs(node.left,target-node.val)+dfs(node.right,target-node.val)
        
        if not root: return 0
        return dfs(root,sum)+self.pathSum(root.left,sum)+self.pathSum(root.right,sum)

 尤其注意是dfs(root,sum) 而不是dfs+dfs+dfs！容易出错！

 

另外解法就是前缀树的解法，前缀树记录sum，比如：

root = [10,sum = 8

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

到节点3的前缀sum就是{10:1,15:1,18:1}，则包括节点3的满足sum=8的路径就是{10,3}这条通路减去{10}这条通路，两个前缀相减，更深的前缀-浅的前缀就是中间路路径，18-10=8满足条件。

代码：

# DeFinition for a binary tree node.
# class TreeNode(object):
#     def __init__(self,sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int        
        """
        def traverse(node,total,pre_sum):
            if not node:
                return 0
            total += node.val
            ans = pre_sum.get(total-sum,0)
            pre_sum[total] = pre_sum.get(total,0)+1
            ans += traverse(node.left,pre_sum) + traverse(node.right,pre_sum)
            pre_sum[total] -= 1
            return ans
        
        return traverse(root,{0:1})