113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

题目含义：找到所有从根节点到叶子节点的路径，使得路径上节点总和等于给定值num

 1     private List<List<Integer>> paths = new ArrayList<>();
 2     private void find(TreeNode root, List<Integer> values, int currentSum, int targetSum) {
 3         if (root == null) return;
 4         values.add(root.val);
 5         currentSum += root.val;
 6 
 7         if (root.left != null && root.right != null) {
 8             find(root.left, values, currentSum, targetSum);
 9             find(root.right, values, currentSum, targetSum);
10         } else if (root.left != null) find(root.left, values, currentSum, targetSum);
11         else if (root.right != null) find(root.right, values, currentSum, targetSum);
12         else if (currentSum == targetSum) //只能是叶子节点，才开始比较值
13         {
14             List<Integer> path = new ArrayList<>(values.size());
15             path.addAll(values);
16             paths.add(path);
17         }
18         values.remove(values.size() - 1);//找到了叶子节点，但是总和不等于目标值，将该叶子节点从values中删除，保证还能使用其上面的values集合来累加其它叶子节点
19     }
20     
21     public List<List<Integer>> pathSum(TreeNode root, int sum) {
22         find(root, new ArrayList<>(), 0, sum);
23         return paths;        
24     }