C++ Program to Print only Odd Numbered Levels of a Tree

นี่คือโปรแกรม C++ เพื่อพิมพ์เฉพาะ Odd Numbered Levels (ระดับคี่) ของ Tree

Algorithm (อัลกอริทึม)

โครงสร้างและฟังก์ชันพร้อม pseudocode:

Begin
    Declare nod as a structure.
        Declare d of integer datatype.
        Declare a pointer l against struct nod.
        Declare a pointer r 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

ตัวอย่าง

#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;
}

ผลลัพธ์

Odd numbered Level Order traversal of binary tree is
7 3 5 2 1