给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。
说明: 叶子节点是指没有子节点的节点。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> L = new ArrayList<List<Integer>>();
List<Integer> list = new ArrayList<Integer>();
if (root == null) {
return L;
}
list.add(root.val);
path(root, sum, list, L, root.val);
return L;
}
public static void path(TreeNode root, int sum, List<Integer> list, List<List<Integer>> L, int cur) {
if (root.left == null && root.right == null) {
if (cur == sum) {
L.add(new ArrayList(list));
return;
}
}
if (root.left != null) {
list.add(root.left.val);
path(root.left, sum, list, L, cur + root.left.val);
list.remove(list.size() - 1);
}
if (root.right != null) {
list.add(root.right.val);
path(root.right, sum, list, L, cur + root.right.val);
list.remove(list.size() - 1);
}
}
}