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);
}