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]
]

與112題方法類似 http://www.lxweimin.com/p/faf516d9914d

Solution1：pre-order DFS(backtracking)

思路： 遞歸pre-order dfs來downwards累積cur_path (backtracking remove回溯)
Time Complexity: O(N) Space Complexity: O(N) 遞歸緩存

Solution2：Interative stack 法二, 紀錄sum和cur_path 并 step back

則不用存sum進stack
Time Complexity: O(N) Space Complexity: O(N)

Solution3：backtracking 套路解法 Round1

Solution1 Code：

class Solution1 {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> result  = new LinkedList<List<Integer>>();
        List<Integer> cur_result  = new LinkedList<Integer>();
        dfs_path(result, cur_result, root, sum);
        return result;
    }
    private void dfs_path(List<List<Integer>> result, List<Integer> cur_result, TreeNode root, int sum) {
        if(root == null) return;
        
        cur_result.add(new Integer(root.val));
        if(root.left == null && root.right == null && sum == root.val) {
            result.add(new LinkedList(cur_result));
            cur_result.remove(cur_result.size() - 1); //remove last one for steping back (backtracking)
            return;
        }
            
        dfs_path(result, cur_result, root.left, sum - root.val);
        dfs_path(result, cur_result, root.right, sum - root.val);
        cur_result.remove(cur_result.size() - 1);
    }
}

Solution2 Code：

class Solution2 {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> cur_path = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        
        TreeNode cur = root;
        while(!stack.isEmpty() || cur != null) {
            while(cur != null) {
                stack.push(cur);
                
                // process
                cur_path.add(cur.val);
                sum -= cur.val;
                if(cur.left == null && cur.right == null && sum == 0) {
                    result.add(new ArrayList<Integer>(cur_path));
                }
                
                cur = cur.left;
            } 
            
            // pop and sum_step_back togeter, but
            // only do that after finishing processing its right_siblings,
            // otherwise when visiting right_siblings, its root will be wrongly sum_step_back.
            while (!stack.isEmpty() && stack.peek().right == cur) {
                cur = stack.pop();
                sum += cur.val;  // sum: step back
                cur_path.remove(cur_path.size() - 1);
                
            }
            cur = stack.isEmpty() ? null : stack.peek().right;
        }
        return result;
    }
}

Solution3 Round1 Code：

class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> result  = new LinkedList<List<Integer>>();
        if(root == null) return result;
        
        List<Integer> cur_result = new LinkedList<Integer>();
        cur_result.add(root.val);
        
        dfs_path(result, cur_result, root, sum - root.val);
        return result;
    }
    private void dfs_path(List<List<Integer>> result, List<Integer> cur_result, TreeNode root, int remain) {
        if(root == null) return;
        
        if(root.left == null && root.right == null && remain == 0) {
            result.add(new LinkedList(cur_result));
            return;
        }
        
        if(root.left != null) {
            cur_result.add(root.left.val);
            dfs_path(result, cur_result, root.left, remain - root.left.val);
            cur_result.remove(cur_result.size() - 1); // steping back
        }
        if(root.right != null) {
            cur_result.add(root.right.val);
            dfs_path(result, cur_result, root.right, remain - root.right.val);
            cur_result.remove(cur_result.size() - 1); // steping back
        }
    }
}

Solution3 Round2 Code：

class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> cur_res = new ArrayList<>();
        
        if(root == null) return result;
        
        dfsCheck(root, sum, cur_res, result);
        return result;    
    }
    
    private void dfsCheck(TreeNode root, int sum_left, List<Integer> cur_res, List<List<Integer>> result) {
        if(root == null) return;
        
        if(root.left == null && root.right == null && root.val == sum_left) {
            cur_res.add(root.val);
            result.add(new ArrayList<>(cur_res));
            cur_res.remove(cur_res.size() - 1);
            return;
        }
        
        cur_res.add(root.val);
        dfsCheck(root.left, sum_left - root.val, cur_res, result);
        dfsCheck(root.right, sum_left - root.val, cur_res, result);
        cur_res.remove(cur_res.size() - 1);
    }
}