LeetCode 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

	public int pathSum(TreeNode root, int sum) {
		return dfspathSum(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
	}

	public int dfspathSum(TreeNode root, int sum) {
		int result = 0;
		if (root == null) return result;
		if (sum == root.val) result++;
		result += dfspathSum(root.left, sum - root.val);
		result += dfspathSum(root.right, sum - root.val);
		return result;
	}

下面的更有效率，19ms，而上面的38ms

	int count = 0;

	public int pathSum2(TreeNode root, int sum) {
		int n = findDepth(root);
		int[] path = new int[n];
		findSum(root, sum, path, 0);
		return count;
	}

	int findDepth(TreeNode root) {
		if (root == null) return 0;
		return Math.max(findDepth(root.left), findDepth(root.right)) + 1;
	}

	void findSum(TreeNode root, int sum, int[] path, int level) {
		if (root == null) return;
		path[level] = root.val;
		int total = 0;
		for (int i = level; i >= 0; i--) {
			total += path[i];
			if (total == sum) {
				count = count + 1;
			}
		}
		findSum(root.left, sum, path, level + 1);
		findSum(root.right, sum, path, level + 1);
		path[level] = Integer.MIN_VALUE;
	}