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 = 8 Output: 3 Explanation: 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 = 22 Output: 3
Constraints:
- The number of nodes in the tree is in the range
[0, 1000]. -109 <= Node.val <= 109-1000 <= targetSum <= 1000
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def pathSum(self, root: TreeNode, targetSum: int) -> int:
preSum = defaultdict(int)
preSum[0] = 1
def dfs(node: TreeNode, cur: int) -> int:
if not node:
return 0
cur += node.val
ret = preSum[cur - targetSum]
preSum[cur] += 1
ret += dfs(node.left, cur)
ret += dfs(node.right, cur)
preSum[cur] -= 1
return ret
return dfs(root, 0)/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private final Map<Integer, Integer> preSum = new HashMap<>();
public int pathSum(TreeNode root, int targetSum) {
preSum.put(0, 1);
return dfs(root, 0, targetSum);
}
private int dfs(TreeNode node, int cur, int targetSum) {
if (node == null) {
return 0;
}
cur += node.val;
int ret = preSum.getOrDefault(cur - targetSum, 0);
preSum.merge(cur, 1, Integer::sum);
ret += dfs(node.left, cur, targetSum);
ret += dfs(node.right, cur, targetSum);
preSum.merge(cur, -1, Integer::sum);
return ret;
}
}/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func pathSum(root *TreeNode, targetSum int) int {
preSum := make(map[int]int)
preSum[0] = 1
var dfs func(*TreeNode, int) int
dfs = func(node *TreeNode, cur int) int {
if node == nil {
return 0
}
cur += node.Val
ret := preSum[cur-targetSum]
preSum[cur]++
ret += dfs(node.Left, cur)
ret += dfs(node.Right, cur)
preSum[cur]--
return ret
}
return dfs(root, 0)
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int pathSum(TreeNode* root, int targetSum) {
unordered_map<int, int> preSum;
preSum[0] = 1;
function<int(TreeNode*, int)> dfs = [&](TreeNode* node, int cur) {
if (node == nullptr) {
return 0;
}
cur += node->val;
int ret = preSum[cur - targetSum];
++preSum[cur];
ret += dfs(node->left, cur);
ret += dfs(node->right, cur);
--preSum[cur];
return ret;
};
return dfs(root, 0);
}
};