動機
原來是prefix sum的靈壓!!
Problem
Given the root of a binary tree and an integer targetSum, return the number of paths where the sum of the values along the path equals targetSum.
The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).
Example 1:
Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8Output: 3Explanation: The paths that sum to 8 are shown.
Example 2:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22Output: 3
Constraints:
- The number of nodes in the tree is in the range
[0, 1000]. -109 <= Node.val <= 109-1000 <= targetSum <= 1000
Sol
最簡單一個走一個找
class Solution:
def pathSum(self, root: TreeNode, val: int) -> int:
def count(r,acc):
if not r:
return 0
elif acc+r.val == val:
ret = 1
else:
ret = 0
ret += count(r.left,acc+r.val)+count(r.right,acc+r.val)
return ret
def dfs(r):
if not r:
return 0
else:
return count(r,0)+dfs(r.left)+dfs(r.right)
return dfs(root)
或是善用prefix sum，不過這是tree，所以要在遞迴離開之前要把原本的sum註銷掉
class Solution:
def pathSum(self, root: TreeNode, targetSum: int) -> int:
sums = defaultdict(int)
sums[0] = 1
ret = 0
def dfs(r,cnt):
nonlocal ret
if not r:
return
else:
cnt += r.val
ret += sums[cnt - targetSum] # targetSum = sums[a:b] = sums[b](cnt) - sums[a]
sums[cnt] += 1
dfs(r.left, cnt)
dfs(r.right, cnt)
sums[cnt] -= 1 # 離開之前要把原本的sum註銷掉
dfs(root,0)
return ret