Leetcode_437

题目描述

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

题解

分类解，包括root节点的和不包括root节点的

public int pathSum(TreeNode root, int sum) {

        if(root==null)
            return 0;
        return findPath(root,sum)+pathSum(root.left,sum)+pathSum(root.right,sum);

    }
    private int findPath(TreeNode node, int sum)
    {
        if(node == null)
            return 0;
        return (node.val==sum ? 1:0) + findPath(node.left,sum-node.val) + findPath(node.right, sum-node.val);
    }