Medium
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:
[0, 1000]
.-109 <= Node.val <= 109
-1000 <= targetSum <= 1000
function pathSum(root: TreeNode | null, targetSum: number): number {
let count = 0
let map = new Map<number, number>()
function dfs(node: TreeNode | null, currentSum: number): void {
if (!node) {
return
}
currentSum += node.val
if (currentSum === targetSum) {
count++
}
count += map.get(currentSum - targetSum) ?? 0
map.set(currentSum, map.get(currentSum) + 1 || 1)
dfs(node?.left, currentSum)
dfs(node?.right, currentSum)
//remove from hashmap
map.set(currentSum, map.get(currentSum) - 1)
if (map.get(currentSum) === 0) {
map.delete(currentSum)
}
}
dfs(root, 0)
return count
}
export { pathSum }