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

给定一个二叉树，求从某一节点开始的路径的和等于给定值，不必从根节点开始，可从二叉树的任意一个节点开始，节点值有正有负。

解法：递归。

Python：

class Solution(object):
def pathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: int
"""
def pathSumHelper(root, curr, sum, lookup):
if root is None:
return 0
curr += root.val
result = lookup[curr-sum] if curr-sum in lookup else 0
lookup[curr] += 1
result += pathSumHelper(root.left, curr, sum, lookup) + \
pathSumHelper(root.right, curr, sum, lookup)
lookup[curr] -= 1
if lookup[curr] == 0:
del lookup[curr]
return result

lookup = collections.defaultdict(int)
lookup[0] = 1
return pathSumHelper(root, 0, sum, lookup)

Python:

class Solution2(object):
def pathSum(self, root, sum):

def pathSumHelper(root, prev, sum):
if root is None:
return 0

curr = prev + root.val;
return int(curr == sum) + \
pathSumHelper(root.left, curr, sum) + \
pathSumHelper(root.right, curr, sum)

if root is None:
return 0

return pathSumHelper(root, 0, sum) + \
self.pathSum(root.left, sum) + \
self.pathSum(root.right, sum)

C++:

class Solution {
public:
int pathSum(TreeNode* root, int sum) {
unordered_map<int, int> m;
m[0] = 1;
return helper(root, sum, 0, m);
}
int helper(TreeNode* node, int sum, int curSum, unordered_map<int, int>& m) {
if (!node) return 0;
curSum += node->val;
int res = m[curSum - sum];
++m[curSum];
res += helper(node->left, sum, curSum, m) + helper(node->right, sum, curSum, m);
--m[curSum];
return res;
}
};　　

C++:

class Solution {
public:
int pathSum(TreeNode* root, int sum) {
if (!root) return 0;
return sumUp(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
}
int sumUp(TreeNode* node, int pre, int& sum) {
if (!node) return 0;
int cur = pre + node->val;
return (cur == sum) + sumUp(node->left, cur, sum) + sumUp(node->right, cur, sum);
}
};

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