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   1Return 3. The paths that sum to 8 are:1.  5 -> 32.  5 -> 2 -> 13. -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;}