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

/**
 * 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) {
        HashMap<Integer, Integer> sums = new HashMap<Integer, Integer>();
        sums.put(0, 1);
        int[] res = new int[1];
        process(root, res, 0, sum, sums);
        return res[0];
    }
    private void process(TreeNode node, int[] res, int currSum, int sum, HashMap<Integer, Integer> sums) {
        if (node == null) {
            return;
        }
        int newSum = currSum + node.val;
        if (sums.containsKey(newSum - sum)) {
            res[0] += sums.get(newSum - sum);
        }
        if (!sums.containsKey(newSum)) {
            sums.put(newSum, 0);
        }
        sums.put(newSum, sums.get(newSum) + 1);
        process(node.left, res, newSum, sum, sums);
        process(node.right, res, newSum, sum, sums);
        sums.put(newSum, sums.get(newSum) - 1);
    }
}

Solution 1: Recursion¶

Time complexity: O(n^2) Space complexity: O(n)

// Running time: 21 ms
class Solution {
public:
  int pathSum(TreeNode* root, int sum) {
    if (!root) return 0;
    return numberOfPaths(root, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
  }
private:
  int numberOfPaths(TreeNode* root, int left) {
    if (!root) return 0;
    left -= root->val;
    return (left == 0 ? 1 : 0) + numberOfPaths(root->left, left) + numberOfPaths(root->right, left);
  }
};

Solution 2: Running Prefix Sum¶

Same idea to 560. Subarray Sum Equals K

Time complexity: O(n) Space complexity: O(h)

// Running time: 13 ms (beats 97.74%)
public:
  int pathSum(TreeNode* root, int sum) {
    ans_ = 0;
    sums_ = {{0, 1}};
    pathSum(root, 0, sum);
    return ans_;
  }
private:
  int ans_;
  unordered_map<int, int> sums_;

  void pathSum(TreeNode* root, int cur, int sum) {
    if (!root) return;
    cur += root->val;
    ans_ += sums_[cur - sum];
    ++sums_[cur];
    pathSum(root->left, cur, sum);
    pathSum(root->right, cur, sum);
    --sums_[cur];
  }
};