[Leetcode] 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]
]

DFS的模板题，在叶子节点的时候进行条件的终止和判断，如果输入是一个null那么直接返回作为corner case，需要注意一下这种情况。

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<Integer>  list = new ArrayList<Integer>();
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        dfs(root, sum, list, result);
        
        return result;
    }
    
    private void dfs(TreeNode root, int sum, List<Integer> list, List<List<Integer>> result) {
        if(root == null) return; // invalid case
        list.add(root.val);
        if(root.left == null && root.right == null) { // left node
            if(sum-root.val == 0) {
                result.add(new ArrayList<Integer>(list));
            }
        }
        else if(root.left == null && root.right != null) {
            dfs(root.right, sum - root.val, list, result);
        }
        else if(root.left != null && root.right == null) {
            dfs(root.left, sum - root.val, list, result);
        }
        else {
            dfs(root.left, sum - root.val, list, result);
            dfs(root.right, sum - root.val, list, result);
        }
        list.remove(list.size() - 1);
    }