[LC] 437. 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

Solution 1: 
Time: O(N^2)
Space: O(Height)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> int:
        if root is None:
            return 0
        cur = self.helper(root, sum) 
        left = self.pathSum(root.left, sum) 
        right = self.pathSum(root.right, sum)
        return cur + left + right
    
    def helper(self, root, sum):
        if root is None:
            return 0
        left = self.helper(root.left, sum - root.val)
        right = self.helper(root.right, sum - root.val)
        if sum == root.val:
            return 1 + left + right
        return left + right

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int count = 0;
    public int pathSum(TreeNode root, int sum) {
        List<Integer> list = new ArrayList<>();
        helper(root, list, sum);
        return count;
    }
    
    private void helper(TreeNode root, List<Integer> list, int sum) {
        if (root == null) {
            return;
        }
        list.add(root.val);
        int num = 0;
        for (int i = list.size() - 1; i >= 0; i--) {
            num += list.get(i);
            if (num == sum) {
                count += 1;
            }
        }
        helper(root.left, list, sum);
        helper(root.right, list, sum);
        list.remove(list.size() - 1);
    }
}

Solution 2:

Time: O(N)

Space: O(N)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> int:
        if root is None:
            return 0
        self.res = 0
        # initilized with 0 as prefix instead of empty b/c need to conver path coming from root
        my_dict = {0 : 1}
        self.helper(root, sum, 0, my_dict) 
        return self.res
    
    def helper(self, root, sum, cur_sum, my_dict):
        if root is None:
            return
        cur_sum += root.val
        reminder = cur_sum - sum
        # use reminder as key
        if reminder in my_dict: 
            self.res += my_dict[reminder]
        my_dict[cur_sum] = my_dict.get(cur_sum, 0) + 1
        # pass cur_sum to children
        left = self.helper(root.left, sum, cur_sum, my_dict)
        right = self.helper(root.right, sum, cur_sum, my_dict)
        my_dict[cur_sum] -= 1

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int count = 0;
    public int pathSum(TreeNode root, int sum) {
        Map<Integer, Integer> map = new HashMap<>();
        // check path from root
        map.put(0, 1);
        helper(root, 0, sum, map);
        return count;
    }
    
    private void helper(TreeNode root, int curSum, int sum, Map<Integer, Integer> map) {
        if (root == null) {
            return;
        }
        int tmpSum = curSum + root.val;
        if (map.containsKey(tmpSum - sum)) {
            count += map.get(tmpSum - sum);
        }
        map.put(tmpSum, map.getOrDefault(tmpSum, 0) + 1);
        helper(root.left, tmpSum, sum, map);
        helper(root.right, tmpSum, sum, map);
        map.put(tmpSum, map.get(tmpSum) - 1);
    }
}