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

    private List<List<Integer>> paths = new ArrayList<>();
    private void find(TreeNode root, List<Integer> values, int currentSum, int targetSum) {
        if (root == null) return;
        values.add(root.val);
        currentSum += root.val;

        if (root.left != null && root.right != null) {
            find(root.left, values, currentSum, targetSum);
            find(root.right, values, currentSum, targetSum);
        } else if (root.left != null) find(root.left, values, currentSum, targetSum);
        else if (root.right != null) find(root.right, values, currentSum, targetSum);
        else if (currentSum == targetSum) //只能是叶子节点，才开始比较值
        {
            List<Integer> path = new ArrayList<>(values.size());
            path.addAll(values);
            paths.add(path);
        }
        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
    }
    
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        find(root, new ArrayList<>(), 0, sum);
        return paths;        
    }