# Path Sum III

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
``````

## Analysis

DFS

Traversing, two level of DFS

Similar to subarray sum:

https://www.jiuzhang.com/solution/path-sum-iii/

https://leetcode.com/problems/path-sum-iii/discuss/91878/17-ms-O(n)-java-Prefix-sum-method

``````具体操作:

backtrack:

``````

## Solution

DFS - Intuitive Implementation

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int pathSum(TreeNode root, int sum) {
if (root == null) {
return 0;
}
return pathHelper(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
}
int pathHelper(TreeNode root, int sum) {
if (root == null) return 0;
return (root.val == sum ? 1 : 0) +
pathHelper(root.left, sum - root.val) + pathHelper(root.right, sum - root.val);
}
}
``````

(Better) Prefix Sum

``````    public int pathSum(TreeNode root, int sum) {
HashMap<Integer, Integer> preSum = new HashMap();
preSum.put(0,1);
return helper(root, 0, sum, preSum);
}

public int helper(TreeNode root, int currSum, int target, HashMap<Integer, Integer> preSum) {
if (root == null) {
return 0;
}

currSum += root.val;
int res = preSum.getOrDefault(currSum - target, 0);
preSum.put(currSum, preSum.getOrDefault(currSum, 0) + 1);

res += helper(root.left, currSum, target, preSum) + helper(root.right, currSum, target, preSum);
preSum.put(currSum, preSum.get(currSum) - 1);
return res;
}
``````