C ++程序仅打印树的奇数级
这是一个仅打印树的奇数级的C++程序。
算法
具有伪代码的结构和功能：
Begin
Declare nod as a structure.
Declare d of integer datatype.
Declare a pointer l against struct nod.
Declare a pointer l against struct nod.
Call function struct nod* newNod(int d).
Declare struct nod* newNod(int d) function.
Declare a pointer node against struct node.
Initialize node = (struct nod*) malloc(sizeof(struct nod)).
node->d = d
node->l = NULL
node->r = NULL
return node.
Call function printLevel(struct nod* root, int lvl).
Declare function printLevel(struct nod* root, int lvl).
if (root == NULL) then
return
if (lvl == 1) then
print the values of root->d
else if (lvl > 1)
call printLevel(root->l, lvl - 1)
printLevel(root->r, lvl - 1)
Call function height(struct nod* node).
Declare function height(struct nod* node) to compute the height of tree.
if (node == NULL) then
return 0
else
int lhght = height(node->l);
int rhght = height(node->r);
if (lhght > rhght) then
return (lhght + 1)
else
return (rhght + 1)
Declare function printLevelOrder(struct nod* root).
declare h of the integer datatype.
initialize h = height(root).
declare i of the integer datatype.
for (i = 1; i <= h; i+=2)
call function printLevel(root, i).
insert values in the tree.
Print “Odd numbered Level Order traversal of binary tree is”.
Call function printLevelOrder(root).
End示例
#include <iostream>
#include<stdlib.h>
using namespace std;
struct nod {
int d;
struct nod* l;
struct nod* r;
};
struct nod* newNod(int d);
struct nod* newNod(int d) {
struct nod* node = (struct nod*) malloc(sizeof(struct nod));
node->d = d;
node->l = NULL;
node->r = NULL;
return (node);
}
void printLevel(struct nod* root, int lvl);
void printLevel(struct nod* root, int lvl) {
if (root == NULL)
return;
if (lvl == 1)
printf("%d ", root->d);
else if (lvl > 1) {
printLevel(root->l, lvl - 1);
printLevel(root->r, lvl - 1);
}
}
int height(struct nod* node);
int height(struct nod* node) {
if (node == NULL)
return 0;
else {
int lhght = height(node->l);
int rhght = height(node->r);
if (lhght > rhght)
return (lhght + 1);
else
return (rhght + 1);
}
}
void printLevelOrder(struct nod* root) {
int h = height(root);
int i;
for (i = 1; i <= h; i+=2)
printLevel(root, i);
}
int main() {
struct nod *root = newNod(7);
root->l = newNod(6);
root->r = newNod(4);
root->l->l = newNod(3);
root->l->r = newNod(5);
root->r->l = newNod(2);
root->r->r = newNod(1);
cout<<"Odd numbered Level Order traversal of binary tree is \n";
printLevelOrder(root);
return 0;
}输出结果
Odd numbered Level Order traversal of binary tree is 7 3 5 2 1