我有一项艰巨的任务,需要你的帮助。
我需要按照以下规则打印二叉树: 不使用矩阵逐级打印; 必须从根打印,你永远不应该在打印后编辑行; 该数字不能与任何其他列相同。 格式如下:
|----10----|
|--13--| 1---|
15 11 0
它不是AVL树。必须在任何大小的树上工作。
这是我到目前为止所做的:
public String printTree() {
if (getAltura() == -1) { //See if the tree is empty
return "Arvore Vazia";
}
if (getAltura() == 0) { //Check if only have one node in the tree;
return raiz.chave + "";
}
return printTree0();
}
private String printTree0() {
String arvore = ""; //String with the binary tree
//String linha = "";
int espaco = 0; //That was what I try to use to put the number under the "|" character
//int altura = 0;
Queue<Nodo> q = new LinkedList<>();
for (int i = 0; i <= getAltura(); i++) {
q.addAll(getNivel(i));
}
while (!q.isEmpty()) {
Nodo n = q.remove();
if (n.direito == null && n.esquerdo == null) {
for (int i = 0; i < espaco; i++) {
arvore += " ";
}
}
if (n.esquerdo != null) { //Check if this node has a left son.
int subarvores = tamanhoSubarvores(n.esquerdo); //Do the math to see how many ASCII character are need to put in this side of the tree.
for (int i = 0; i < subarvores; i++) {
arvore += " ";
espaco++;
}
arvore += "|";
for (int i = 0; i < subarvores; i++) {
arvore += "-";
}
}
arvore += n.chave;
if (n.direito != null) { //Check if this node has a right son.
int subarvores = tamanhoSubarvores(n.direito); //Do the math to see how many ASCII character are need to put in this side of the tree.
for (int i = 0; i < subarvores; i++) {
arvore += "-";
}
arvore += "|";
for (int i = 0; i < subarvores; i++) {
arvore += " ";
}
}
arvore += "\n";
}
return arvore;
}
private int tamanhoSubarvores(Nodo nodo) {
int size = 0;
for (Nodo n : getNivel(nodo.altura, nodo)) {
size += Integer.toString(n.chave).length();
}
return size;
}
谢谢。
答案 0 :(得分:1)
您要做的事情称为Breadth First Search。假设在添加或删除元素期间AVL tree
仅与普通Binary Search Tree
不同,则应用上面wiki链接中列出的BF-Search的算法逻辑。