原文:
给定一个二叉树，任务是从左到右遍历给定二叉树的每一级，并打印在某一级遇到的每个备选项。
示例:
输入:
输出:t2 1 2 3 9 5 7
输入:
输出:t2】71 88 4 6 8 10 13
方法:对给定的二叉树进行遍历即可解决问题。按照以下步骤解决问题:
下面是上述方法的实现:
c
// c program to implement
// the above approach
#include
using namespace std;
// structure of a node
struct node {
int data;
node* left;
node* right;
node(int val)
{
data = val;
left = right = null;
}
};
// print alternate nodes of
// a binary tree
void printalternate(node* root)
{
// store nodes of each level
queue q;
q.push(root);
while (!q.empty()) {
// store count of nodes
// of current level
int n = q.size();
// print alternate nodes of
// the current level
for (int i = 0; i < n; i ) {
node* temp = q.front();
q.pop();
if (i % 2 == 0) {
cout << temp->data << " ";
}
// if left child exists
if (temp->left) {
// store left child
q.push(temp->left);
}
// if right child exists
if (temp->right) {
// store right child
q.push(temp->right);
}
}
cout << endl;
}
}
// driver code
int main()
{
node* root;
// create a tree
root = new node(71);
root->left = new node(88);
root->right = new node(99);
root->left->left = new node(4);
root->left->right = new node(5);
root->right->left = new node(6);
root->right->right = new node(7);
root->left->left->left = new node(8);
root->left->left->right = new node(9);
root->left->right->left = new node(10);
root->left->right->right = new node(11);
root->right->left->right = new node(13);
root->right->right->left = new node(14);
printalternate(root);
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to implement
// the above approach
import java.util.*;
class gfg{
// structure of a node
static class node
{
int data;
node left;
node right;
node(int val)
{
data = val;
left = right = null;
}
};
// print alternate nodes of
// a binary tree
static void printalternate(node root)
{
// store nodes of each level
queue q = new linkedlist<>();
q.add(root);
while (!q.isempty())
{
// store count of nodes
// of current level
int n = q.size();
// print alternate nodes of
// the current level
for (int i = 0; i < n; i )
{
node temp = q.peek();
q.remove();
if (i % 2 == 0)
{
system.out.print(temp.data " ");
}
// if left child exists
if (temp.left!=null)
{
// store left child
q.add(temp.left);
}
// if right child exists
if (temp.right!=null)
{
// store right child
q.add(temp.right);
}
}
system.out.println();
}
}
// driver code
public static void main(string[] args)
{
node root;
// create a tree
root = new node(71);
root.left = new node(88);
root.right = new node(99);
root.left.left = new node(4);
root.left.right = new node(5);
root.right.left = new node(6);
root.right.right = new node(7);
root.left.left.left = new node(8);
root.left.left.right = new node(9);
root.left.right.left = new node(10);
root.left.right.right = new node(11);
root.right.left.right = new node(13);
root.right.right.left = new node(14);
printalternate(root);
}
}
// this code is contributed by 29ajaykumar
python 3
# python3 program to implement
# the above approach
# structure of a node
class newnode:
def __init__(self, val):
self.data = val
self.left = none
self.right = none
# print alternate nodes of
# a binary tree
def printalternate(root):
# store nodes of each level
q = []
q.append(root)
while (len(q)):
# store count of nodes
# of current level
n = len(q)
# print alternate nodes of
# the current level
for i in range(n):
temp = q[0]
q.remove(q[0])
if (i % 2 == 0):
print(temp.data, end = " ")
# if left child exists
if (temp.left):
# store left child
q.append(temp.left)
# if right child exists
if (temp.right):
# store right child
q.append(temp.right)
print("\n", end = "")
# driver code
if __name__ == '__main__':
# create a tree
root = newnode(71)
root.left = newnode(88)
root.right = newnode(99)
root.left.left = newnode(4)
root.left.right = newnode(5)
root.right.left = newnode(6)
root.right.right = newnode(7)
root.left.left.left = newnode(8)
root.left.left.right = newnode(9)
root.left.right.left = newnode(10)
root.left.right.right = newnode(11)
root.right.left.right = newnode(13)
root.right.right.left = newnode(14)
printalternate(root)
# this code is contributed by ipg2016107
c
// c# program to implement
// the above approach
using system;
using system.collections.generic;
class gfg{
// structure of a node
class node
{
public int data;
public node left;
public node right;
public node(int val)
{
data = val;
left = right = null;
}
};
// print alternate nodes of
// a binary tree
static void printalternate(node root)
{
// store nodes of each level
queue q = new queue();
q.enqueue(root);
while (q.count != 0)
{
// store count of nodes
// of current level
int n = q.count;
// print alternate nodes of
// the current level
for (int i = 0; i < n; i )
{
node temp = q.peek();
q.dequeue();
if (i % 2 == 0)
{
console.write(temp.data " ");
}
// if left child exists
if (temp.left!=null)
{
// store left child
q.enqueue(temp.left);
}
// if right child exists
if (temp.right!=null)
{
// store right child
q.enqueue(temp.right);
}
}
console.writeline();
}
}
// driver code
public static void main(string[] args)
{
node root;
// create a tree
root = new node(71);
root.left = new node(88);
root.right = new node(99);
root.left.left = new node(4);
root.left.right = new node(5);
root.right.left = new node(6);
root.right.right = new node(7);
root.left.left.left = new node(8);
root.left.left.right = new node(9);
root.left.right.left = new node(10);
root.left.right.right = new node(11);
root.right.left.right = new node(13);
root.right.right.left = new node(14);
printalternate(root);
}
}
// this code is contributed by 29ajaykumar
java 描述语言
output:
71
88
4 6
8 10 13
时间复杂度:o(n) t5辅助空间:** o(n)
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