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

O(n^2)
每个点作为起点，遍历结果

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

O(nlogn)  记录<preSum,Count>并且对每个点在preSum Map里找(sum-target)的值，sum是从root下来的path的总和，如果对于当前位置而言，(sum-target)出现过那么sum-前面对应位置的sum就是target，也就是说这两个位置之间的和是target。 对于preSum记录count 就能找到多个位置满足(当前sum-target == preSum)。 思路是如果这个点是end point 的话，count要加。