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给定一棵二叉树和两个级别数字 ” 低 ” 和 ” 高 ”, 从低级别到高级别打印节点。
For example consider the binary tree given in below diagram.
Input: Root of below tree, low = 2, high = 4
Output:
8 22
4 12
10 14
一种简略办法
首先编写一个递归函数, 该函数打印给定级别编号的节点。而后在从低到高的循环中调用递归函数。该办法的工夫复杂度为 O(n
2) 咱们能够打印节点，在 O(n) 工夫，应用基于队列的迭代级别程序遍历。这个想法是做简略的基于队列的级别程序遍历。在进行有序遍历时, 在开端增加一个标记节点。每当咱们看到标记节点时, 咱们都会减少级别号。如果级别号介于低和高之间, 则打印节点。
以下是上述想法的实现。
C ++
// A C++ program to print Nodes level by level berween given two levels.
#include <bits/stdc++.h>
using namespace std;
/* A binary tree Node has key, pointer to left and right children */
struct Node
{
int key;
struct Node* left, *right;
};
/* Given a binary tree, print nodes from level number 'low' to level
number 'high'*/
void printLevels(Node* root, int low, int high)
{
queue <Node *> Q;
Node *marker = new Node; // Marker node to indicate end of level
int level = 1; // Initialize level number
// Enqueue the only first level node and marker node for end of level
Q.push(root);
Q.push(marker);
// Simple level order traversal loop
while (Q.empty() == false )
{
// Remove the front item from queue
Node *n = Q.front();
Q.pop();
// Check if end of level is reached
if (n == marker)
{
// print a new line and increment level number
cout << endl;
level++;
// Check if marker node was last node in queue or
// level number is beyond the given upper limit
if (Q.empty() == true || level > high) break ;
// Enqueue the marker for end of next level
Q.push(marker);
// If this is marker, then we don't need print it
// and enqueue its children
continue ;
}
// If level is equal to or greater than given lower level, // print it
if (level >= low)
cout << n->key << " " ;
// Enqueue children of non-marker node
if (n->left != NULL) Q.push(n->left);
if (n->right != NULL) Q.push(n->right);
}
}
/* Helper function that allocates a new Node with the
given key and NULL left and right pointers. */
Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
temp->left = temp->right = NULL;
return (temp);
}
/* Driver program to test above functions*/
int main()
{
// Let us construct the BST shown in the above figure
struct Node *root = newNode(20);
root->left = newNode(8);
root->right = newNode(22);
root->left->left = newNode(4);
root->left->right = newNode(12);
root->left->right->left = newNode(10);
root->left->right->right = newNode(14);
cout << "Level Order traversal between given two levels is" ;
printLevels(root, 2, 3);
return 0;
}Java
// Java program to print Nodes level by level between given two levels
import java.util.LinkedList;
import java.util.Queue;
/* A binary tree Node has key, pointer to left and right children */
class Node
{
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null ;
}
}
class BinaryTree
{
Node root;
/* Given a binary tree, print nodes from level number 'low' to level
number 'high'*/
void printLevels(Node node, int low, int high)
{Queue<Node> Q = new LinkedList<>();
Node marker = new Node(4); // Marker node to indicate end of level
int level = 1 ; // Initialize level number
// Enqueue the only first level node and marker node for end of level
Q.add(node);
Q.add(marker);
// Simple level order traversal loop
while (Q.isEmpty() == false )
{
// Remove the front item from queue
Node n = Q.peek();
Q.remove();
// Check if end of level is reached
if (n == marker)
{
// print a new line and increment level number
System.out.println("");
level++;
// Check if marker node was last node in queue or
// level number is beyond the given upper limit
if (Q.isEmpty() == true || level > high)
break ;
// Enqueue the marker for end of next level
Q.add(marker);
// If this is marker, then we don't need print it
// and enqueue its children
continue ;
}
// If level is equal to or greater than given lower level, // print it
if (level >= low)
System.out.print(n.data + " ");
// Enqueue children of non-marker node
if (n.left != null)
Q.add(n.left);
if (n.right != null)
Q.add(n.right);
}
}
// Driver program to test for above functions
public static void main(String args[])
{BinaryTree tree = new BinaryTree();
tree.root = new Node(20);
tree.root.left = new Node(8);
tree.root.right = new Node(22);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(12);
tree.root.left.right.left = new Node(10);
tree.root.left.right.right = new Node(14);
System.out.print("Level Order traversal between given two levels is");
tree.printLevels(tree.root, 2 , 3);
}
}
// This code has been contributed by Mayank Jaiswalpython
# Python program to print nodes level by level between
# given two levels
# A binary tree node
class Node:
# Constructor tor create a new node
def __init__(self , key):
self .key = key
self .left = None
self .right = None
# Given a binary tree, print nodes form level number 'low'
# to level number 'high'
def printLevels(root, low, high):
Q = []
marker = Node(11114) # Marker node to indicate end of level
level = 1 # Initialize level number
# Enqueue the only first level node and marker node for
# end of level
Q.append(root)
Q.append(marker)
#print Q
# Simple level order traversal loop
while (len (Q) > 0 ):
# Remove the front item from queue
n = Q[0]
Q.pop(0)
#print Q
# Check if end of level is reached
if n = = marker:
# print a new line and increment level number
print
level + = 1
# Check if marker node was last node in queue
# or level nubmer is beyond the given upper limit
if len (Q) = = 0 or level > high:
break
# Enqueue the marker for end of next level
Q.append(marker)
# If this is marker, then we don't need print it
# and enqueue its children
continue
if level > = low:
print n.key, # Enqueue children of non-marker node
if n.left is not None :
Q.append(n.left)
Q.append(n.right)
# Driver program to test the above function
root = Node(20)
root.left = Node(8)
root.right = Node(22)
root.left.left = Node(4)
root.left.right = Node(12)
root.left.right.left = Node(10)
root.left.right.right = Node(14)
print "Level Order Traversal between given two levels is" , printLevels(root, 2 , 3)
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)C
using System;
using System.Collections.Generic;
// c# program to print Nodes level by level between given two levels
/* A binary tree Node has key, pointer to left and right children */
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null ;
}
}
public class BinaryTree
{
public Node root;
/* Given a binary tree, print nodes from level number 'low' to level
number 'high'*/
public virtual void printLevels(Node node, int low, int high)
{LinkedList<Node> Q = new LinkedList<Node>();
Node marker = new Node(4); // Marker node to indicate end of level
int level = 1; // Initialize level number
// Enqueue the only first level node and marker node for end of level
Q.AddLast(node);
Q.AddLast(marker);
// Simple level order traversal loop
while (Q.Count > 0)
{
// Remove the front item from queue
Node n = Q.First.Value;
Q.RemoveFirst();
// Check if end of level is reached
if (n == marker)
{
// print a new line and increment level number
Console.WriteLine("");
level++;
// Check if marker node was last node in queue or
// level number is beyond the given upper limit
if (Q.Count == 0 || level > high)
{break ;}
// Enqueue the marker for end of next level
Q.AddLast(marker);
// If this is marker, then we don't need print it
// and enqueue its children
continue ;
}
// If level is equal to or greater than given lower level, // print it
if (level >= low)
{Console.Write(n.data + " ");
}
// Enqueue children of non-marker node
if (n.left != null)
{Q.AddLast(n.left);
}
if (n.right != null)
{Q.AddLast(n.right);
}
}
}
// Driver program to test for above functions
public static void Main(string [] args)
{BinaryTree tree = new BinaryTree();
tree.root = new Node(20);
tree.root.left = new Node(8);
tree.root.right = new Node(22);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(12);
tree.root.left.right.left = new Node(10);
tree.root.left.right.right = new Node(14);
Console.Write("Level Order traversal between given two levels is");
tree.printLevels(tree.root, 2, 3);
}
}
// This code is contributed by Shrikant13输入如下
Level Order traversal between given two levels is
8 22
4 12上述办法的工夫复杂度为 O(n), 因为它执行简略的层级程序遍历。
如果发现任何不正确的中央, 或者想分享无关上述主题的更多信息, 请发表评论。
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