• Time:O(n)
• Space:O(n)

## C++

``````class Solution {
public:
int pathSum(vector<int>& nums) {
int ans = 0;
vector<vector<int>> tree(4, vector<int>(8, -1));

for (const int num : nums) {
const int d = num / 100 - 1;
const int p = (num % 100) / 10 - 1;
const int v = num % 10;
tree[d][p] = v;
}

dfs(tree, 0, 0, 0, ans);
return ans;
}

private:
void dfs(const vector<vector<int>>& tree, int i, int j, int path, int& ans) {
if (tree[i][j] == -1)
return;
if (i == 3 || max(tree[i + 1][j * 2], tree[i + 1][j * 2 + 1]) == -1) {
ans += path + tree[i][j];
return;
}

dfs(tree, i + 1, j * 2, path + tree[i][j], ans);
dfs(tree, i + 1, j * 2 + 1, path + tree[i][j], ans);
}
};
``````

## JAVA

``````class Solution {
public int pathSum(int[] nums) {
int[][] tree = new int[4][8];
Arrays.stream(tree).forEach(A -> Arrays.fill(A, -1));

for (final int num : nums) {
final int d = num / 100 - 1;
final int p = (num % 100) / 10 - 1;
final int v = num % 10;
tree[d][p] = v;
}

dfs(tree, 0, 0, 0);
return ans;
}

private int ans = 0;

private void dfs(int[][] tree, int i, int j, int path) {
if (tree[i][j] == -1)
return;
if (i == 3 || Math.max(tree[i + 1][j * 2], tree[i + 1][j * 2 + 1]) == -1) {
ans += path + tree[i][j];
return;
}

dfs(tree, i + 1, j * 2, path + tree[i][j]);
dfs(tree, i + 1, j * 2 + 1, path + tree[i][j]);
}
}
``````

## Python

``````class Solution:
def pathSum(self, nums: List[int]) -> int:
ans = 0
tree = [[-1] * 8 for _ in range(4)]

for num in nums:
d = num // 100 - 1
p = (num % 100) // 10 - 1
v = num % 10
tree[d][p] = v

def dfs(i: int, j: int, path: int) -> None:
nonlocal ans
if tree[i][j] == -1:
return
if i == 3 or max(tree[i + 1][j * 2], tree[i + 1][j * 2 + 1]) == -1:
ans += path + tree[i][j]
return

dfs(i + 1, j * 2, path + tree[i][j])
dfs(i + 1, j * 2 + 1, path + tree[i][j])

dfs(0, 0, 0)
return ans
``````