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);
}
};
类似题目：
[LeetCode] 113. Path Sum II 路径和 II