给你一个二叉树的根节点 root
,树中每个节点都存放有一个 0
到 9
之间的数字。
每条从根节点到叶节点的路径都代表一个数字:
1 -> 2 -> 3
表示数字 123
。计算从根节点到叶节点生成的 所有数字之和 。
叶节点 是指没有子节点的节点。
class Solution {
public int sumNumbers(TreeNode root) {
return dfs(root, 0);
}
public int dfs(TreeNode root, int temp) {
if (root == null) {
return 0;
}
int result = temp * 10 + root.val;
if (root.left == null && root.right == null) {
return result;
} else {
return dfs(root.left, result) + dfs(root.right, result);
}
}
}
给你一个 m x n
的矩阵 board
,由若干字符 'X'
和 'O'
,找到所有被 'X'
围绕的区域,并将这些区域里所有的 'O'
用 'X'
填充。
输入:board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
输出:[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
解释:被围绕的区间不会存在于边界上,换句话说,任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上,或不与边界上的 'O' 相连的 'O' 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻,则称它们是“相连”的。
class Solution {
int n;
int m;
public void solve(char[][] board) {
n = board.length;
if (n == 0) {
return;
}
m = board[0].length;
for (int i = 0; i < n; i++) {
dfs(board, i, 0);
dfs(board, i, m - 1);
}
for (int i = 1; i < m - 1; i++) {
dfs(board, 0, i);
dfs(board, n - 1, i);
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (board[i][j] == 'O') {
board[i][j] = 'X';
} else if (board[i][j] == 'M') {
board[i][j] = 'O';
}
}
}
}
public void dfs(char[][] board, int x, int y) {
if (x < 0 || x >= n || y < 0 || y >= m || board[x][y] != 'O') {
return;
}
board[x][y] = 'M';
dfs(board, x - 1, y);
dfs(board, x + 1, y);
dfs(board, x, y - 1);
dfs(board, x, y + 1);
}
}