我需要对二进制搜索树进行顺序遍历,我需要打印所有节点和它们所处的级别,但我想不出这样做的方法。
例如:
如果我有这个bst,输出将是:
4 #1
5 #0
9 #1
7 #2
10 #2
18 #3
到目前为止,这是我得到的:
这是我正在使用的结构:
struct tree {
int number;
tree *izq;
tree *der;
};
typedef struct tree *bst;
这是我正在尝试实现的功能:
void printTree(bst node) {
if (node==NULL) {
return;
}
else {
if (node->left != NULL) {
printTree(node->left);
}
printf("%d", node->number);
if (node->right !=NULL) {
printTree(node->right);
}
}
}
有没有人有任何想法?谢谢 :)
警告
struct tree {
int number;
tree *izq;
tree *der;
};
一定是
struct tree {
int number;
struct tree *izq;
struct tree *der;
};
因为在开始时检查节点为NULL的情况,您可以简化:
void printTree(bst node) {
if (node != NULL) {
printTree(node->izq);
printf("%d", node->number);
printTree(node->der);
}
}
添加级别:
void printTree(bst node, int lvl) {
if (node != NULL) {
printTree(node->izq, lvl + 1);
printf("%d #%d\n", node->number, lvl);
printTree(node->der, lvl + 1);
}
}
并在级别为0的根级别调用
制作完整的计划:
#include <stdio.h>
#include <stdlib.h>
struct tree {
int number;
struct tree *izq;
struct tree *der;
};
typedef struct tree *bst;
void printTree(bst node, int lvl) {
if (node != NULL) {
printTree(node->izq, lvl + 1);
printf("%d #%d\n", node->number, lvl);
printTree(node->der, lvl + 1);
}
}
struct tree * mk(int v, struct tree * l, struct tree * r)
{
struct tree * t = malloc(sizeof(struct tree));
t->number = v;
t->izq = l;
t->der = r;
return t;
}
int main()
{
struct tree * r = mk(5, mk(4, NULL, NULL), mk(9, mk(7, NULL, NULL), mk(10, NULL, mk(18, NULL, NULL))));
printTree(r, 0);
}
编译执行:
pi@raspberrypi:/tmp $ gcc -pedantic -Wall -Wextra c.c
pi@raspberrypi:/tmp $ ./a.out
4 #1
5 #0
7 #2
9 #1
10 #2
18 #3
我用Java实现了这个,但我认为你可以很容易地将它转换为C:
private static void printWithLevels(TreeNode node) {
printWithLevels(node, 0);
}
private static void printWithLevels(TreeNode node, int level) {
if (node == null) return;
System.out.println(node.value + "(" + level + ")");
printWithLevels(node.left, level + 1);
printWithLevels(node.right, level + 1);
}
为了使我的解决方案完整,这是我对TreeNode的简单/快速实现:
private static class TreeNode {
int value;
TreeNode left;
TreeNode right;
TreeNode(int value, TreeNode left, TreeNode right) {
this.value = value;
this.left = left;
this.right = right;
}
}