在 BFS 中创建一个接受用户输入的树
Creating a tree accpting users input in BFS
我很困惑如何在遍历 BFS 时保持树的顶部。我正在接受这样的输入:
5 // total number of nodes in tree (root node will always be 1)
// this number also tells the number of lines that'll be coming up
2 3 // left and right child of root, tree be like : 1
/ \
2 3
4 5 // left and right child of 2 , tree 1
/ \
2 3
/ \
4 5
-1 -1 // left and right child of 3 , that's null
-1 -1 // left and right child of 4
-1 -1 // left and right child of 5
以上述方式继续,用户在 BFS 中输入左右 child。但是我无法理解如何实现这一目标。
我的看法是:
LinkedList<Node> list = new LinkedList<Node>
list.add(root); //initially push root
while(list.peek()){ //while there is node left in linked list
curr = list.poll();
curr.left = new Node(x) // x = whatever is the user input is (using nextInt)
// and if it's -1, i'll make left child null
curr.right = new Node(x) // same goes for this one
ll.add(curr);
}
最后,我需要根节点的引用,不知如何获取?还有,有什么更好的方法可以完成这个任务
希望下面的代码对您有所帮助。
"readTree"函数用于读取树
public static Node readTree(Scanner in) {
Queue<Node> queue = new LinkedList<Node>();
Node root = new Node(1);
queue.add(root);
while (!queue.isEmpty()) {
Node node = queue.poll();
int left = in.nextInt();
int right = in.nextInt();
if (-1 != left) {
node.left = new Node(left);
queue.add(node.left);
}
if (-1 != right) {
node.right = new Node(right);
queue.add(node.right);
}
}
return root;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
in.nextInt(); // number of nodes is not used.
Node result = readTree(in);
}
我很困惑如何在遍历 BFS 时保持树的顶部。我正在接受这样的输入:
5 // total number of nodes in tree (root node will always be 1)
// this number also tells the number of lines that'll be coming up
2 3 // left and right child of root, tree be like : 1
/ \
2 3
4 5 // left and right child of 2 , tree 1
/ \
2 3
/ \
4 5
-1 -1 // left and right child of 3 , that's null
-1 -1 // left and right child of 4
-1 -1 // left and right child of 5
以上述方式继续,用户在 BFS 中输入左右 child。但是我无法理解如何实现这一目标。
我的看法是:
LinkedList<Node> list = new LinkedList<Node>
list.add(root); //initially push root
while(list.peek()){ //while there is node left in linked list
curr = list.poll();
curr.left = new Node(x) // x = whatever is the user input is (using nextInt)
// and if it's -1, i'll make left child null
curr.right = new Node(x) // same goes for this one
ll.add(curr);
}
最后,我需要根节点的引用,不知如何获取?还有,有什么更好的方法可以完成这个任务
希望下面的代码对您有所帮助。
"readTree"函数用于读取树
public static Node readTree(Scanner in) {
Queue<Node> queue = new LinkedList<Node>();
Node root = new Node(1);
queue.add(root);
while (!queue.isEmpty()) {
Node node = queue.poll();
int left = in.nextInt();
int right = in.nextInt();
if (-1 != left) {
node.left = new Node(left);
queue.add(node.left);
}
if (-1 != right) {
node.right = new Node(right);
queue.add(node.right);
}
}
return root;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
in.nextInt(); // number of nodes is not used.
Node result = readTree(in);
}