获取二叉树中节点的级别|迭代方法
原文:https://www . geesforgeks . org/get-level-node-二叉树-iterative-approach/
给定一棵二叉树和一个关键字，编写一个返回关键字级别的函数。 例如，考虑下面的树。如果输入键是 3，那么你的函数应该返回 1。如果输入键是 4，那么你的函数应该返回 3。对于键中不存在的键，您的函数应该返回 0。
这里讨论了这个问题的递归方法
迭代方法讨论如下: 迭代方法是级序树遍历 算法 的修改版本
create a empty queue q
push root and then NULL to q
loop till q is not empty
get the front node into temp node
if it is NULL, it means all nodes of
one level are traversed, so increment
level
else
check if temp data is equal to data
to be searched
if yes then return level
if temp->left is not NULL,
enqueue temp->left
if temp->right is not NULL,
enqueue temp->right
if value not found, then return 0
C++
// CPP program to print level of given node
// in binary tree iterative approach
/* Example binary tree
root is at level 1
20
/ \
10 30
/ \ / \
5 15 25 40
/
12 */
#include <bits/stdc++.h>
using namespace std;
// node of binary tree
struct node {
int data;
node* left;
node* right;
};
// utility function to create
// a new node
node* getnode(int data)
{
node* newnode = new node();
newnode->data = data;
newnode->left = NULL;
newnode->right = NULL;
}
// utility function to return level of given node
int getlevel(node* root, int data)
{
queue<node*> q;
int level = 1;
q.push(root);
// extra NULL is pushed to keep track
// of all the nodes to be pushed before
// level is incremented by 1
q.push(NULL);
while (!q.empty()) {
node* temp = q.front();
q.pop();
if (temp == NULL) {
if (q.front() != NULL) {
q.push(NULL);
}
level += 1;
} else {
if (temp->data == data) {
return level;
}
if (temp->left) {
q.push(temp->left);
}
if (temp->right) {
q.push(temp->right);
}
}
}
return 0;
}
int main()
{
// create a binary tree
node* root = getnode(20);
root->left = getnode(10);
root->right = getnode(30);
root->left->left = getnode(5);
root->left->right = getnode(15);
root->left->right->left = getnode(12);
root->right->left = getnode(25);
root->right->right = getnode(40);
// return level of node
int level = getlevel(root, 30);
(level != 0) ? (cout << "level of node 30 is " << level << endl) :
(cout << "node 30 not found" << endl);
level = getlevel(root, 12);
(level != 0) ? (cout << "level of node 12 is " << level << endl) :
(cout << "node 12 not found" << endl);
level = getlevel(root, 25);
(level != 0) ? (cout << "level of node 25 is " << level << endl) :
(cout << "node 25 not found" << endl);
level = getlevel(root, 27);
(level != 0) ? (cout << "level of node 27 is " << level << endl) :
(cout << "node 27 not found" << endl);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print level of given node
// in binary tree iterative approach
/* Example binary tree
root is at level 1
20
/ \
10 30
/ \ / \
5 15 25 40
/
12 */
import java.io.*;
import java.util.*;
class GFG
{
// node of binary tree
static class node
{
int data;
node left, right;
node(int data)
{
this.data = data;
this.left = this.right = null;
}
}
// utility function to return level of given node
static int getLevel(node root, int data)
{
Queue<node> q = new LinkedList<>();
int level = 1;
q.add(root);
// extra NULL is pushed to keep track
// of all the nodes to be pushed before
// level is incremented by 1
q.add(null);
while (!q.isEmpty())
{
node temp = q.poll();
if (temp == null)
{
if (q.peek() != null)
{
q.add(null);
}
level += 1;
}
else
{
if (temp.data == data)
{
return level;
}
if (temp.left != null)
{
q.add(temp.left);
}
if (temp.right != null)
{
q.add(temp.right);
}
}
}
return 0;
}
// Driver Code
public static void main(String[] args)
{
// create a binary tree
node root = new node(20);
root.left = new node(10);
root.right = new node(30);
root.left.left = new node(5);
root.left.right = new node(15);
root.left.right.left = new node(12);
root.right.left = new node(25);
root.right.right = new node(40);
// return level of node
int level = getLevel(root, 30);
if (level != 0)
System.out.println("level of node 30 is " + level);
else
System.out.println("node 30 not found");
level = getLevel(root, 12);
if (level != 0)
System.out.println("level of node 12 is " + level);
else
System.out.println("node 12 not found");
level = getLevel(root, 25);
if (level != 0)
System.out.println("level of node 25 is " + level);
else
System.out.println("node 25 not found");
level = getLevel(root, 27);
if (level != 0)
System.out.println("level of node 27 is " + level);
else
System.out.println("node 27 not found");
}
}
// This code is contributed by
// sanjeev2552
Python 3
# Python3 program to find closest
# value in Binary search Tree
_MIN = -2147483648
_MAX = 2147483648
# Helper function that allocates a new
# node with the given data and None
# left and right poers.
class getnode:
# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# utility function to return level
# of given node
def getlevel(root, data):
q = []
level = 1
q.append(root)
# extra None is appended to keep track
# of all the nodes to be appended
# before level is incremented by 1
q.append(None)
while (len(q)):
temp = q[0]
q.pop(0)
if (temp == None) :
if len(q) == 0:
return 0
if (q[0] != None):
q.append(None)
level += 1
else :
if (temp.data == data) :
return level
if (temp.left):
q.append(temp.left)
if (temp.right) :
q.append(temp.right)
return 0
# Driver Code
if __name__ == '__main__':
# create a binary tree
root = getnode(20)
root.left = getnode(10)
root.right = getnode(30)
root.left.left = getnode(5)
root.left.right = getnode(15)
root.left.right.left = getnode(12)
root.right.left = getnode(25)
root.right.right = getnode(40)
# return level of node
level = getlevel(root, 30)
if level != 0:
print("level of node 30 is", level)
else:
print("node 30 not found")
level = getlevel(root, 12)
if level != 0:
print("level of node 12 is", level)
else:
print("node 12 not found")
level = getlevel(root, 25)
if level != 0:
print("level of node 25 is", level)
else:
print("node 25 not found")
level = getlevel(root, 27)
if level != 0:
print("level of node 27 is", level)
else:
print("node 27 not found")
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)
C
// C# program to print level of given node
// in binary tree iterative approach
/* Example binary tree
root is at level 1
20
/ \
10 30
/ \ / \
5 15 25 40
/
12 */
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
// node of binary tree
public class node
{
public int data;
public node left, right;
public node(int data)
{
this.data = data;
this.left = this.right = null;
}
}
// utility function to return level of given node
static int getLevel(node root, int data)
{
Queue<node> q = new Queue<node>();
int level = 1;
q.Enqueue(root);
// extra NULL is pushed to keep track
// of all the nodes to be pushed before
// level is incremented by 1
q.Enqueue(null);
while (q.Count > 0)
{
node temp = q.Dequeue();
if (temp == null)
{
if (q.Count > 0)
{
q.Enqueue(null);
}
level += 1;
}
else
{
if (temp.data == data)
{
return level;
}
if (temp.left != null)
{
q.Enqueue(temp.left);
}
if (temp.right != null)
{
q.Enqueue(temp.right);
}
}
}
return 0;
}
// Driver Code
public static void Main(String []args)
{
// create a binary tree
node root = new node(20);
root.left = new node(10);
root.right = new node(30);
root.left.left = new node(5);
root.left.right = new node(15);
root.left.right.left = new node(12);
root.right.left = new node(25);
root.right.right = new node(40);
// return level of node
int level = getLevel(root, 30);
if (level != 0)
Console.WriteLine("level of node 30 is " + level);
else
Console.WriteLine("node 30 not found");
level = getLevel(root, 12);
if (level != 0)
Console.WriteLine("level of node 12 is " + level);
else
Console.WriteLine("node 12 not found");
level = getLevel(root, 25);
if (level != 0)
Console.WriteLine("level of node 25 is " + level);
else
Console.WriteLine("node 25 not found");
level = getLevel(root, 27);
if (level != 0)
Console.WriteLine("level of node 27 is " + level);
else
Console.WriteLine("node 27 not found");
}
}
// This code is contributed by Arnab Kundu
java 描述语言
<script>
// Javascript program to print level of given node
// in binary tree iterative approach
/* Example binary tree
root is at level 1
20
/ \
10 30
/ \ / \
5 15 25 40
/
12 */
class node
{
constructor(data) {
this.left = null;
this.right = null;
this.data = data;
}
}
// utility function to return level of given node
function getLevel(root, data)
{
let q = [];
let level = 1;
q.push(root);
// extra NULL is pushed to keep track
// of all the nodes to be pushed before
// level is incremented by 1
q.push(null);
while (q.length > 0)
{
let temp = q[0];
q.shift()
if (temp == null)
{
if (q[0] != null)
{
q.push(null);
}
level += 1;
}
else
{
if (temp.data == data)
{
return level;
}
if (temp.left != null)
{
q.push(temp.left);
}
if (temp.right != null)
{
q.push(temp.right);
}
}
}
return 0;
}
// create a binary tree
let root = new node(20);
root.left = new node(10);
root.right = new node(30);
root.left.left = new node(5);
root.left.right = new node(15);
root.left.right.left = new node(12);
root.right.left = new node(25);
root.right.right = new node(40);
// return level of node
let level = getLevel(root, 30);
if (level != 0)
document.write("level of node 30 is " + level + "</br>");
else
document.write("node 30 not found" + "</br>");
level = getLevel(root, 12);
if (level != 0)
document.write("level of node 12 is " + level + "</br>");
else
document.write("node 12 not found" + "</br>");
level = getLevel(root, 25);
if (level != 0)
document.write("level of node 25 is " + level + "</br>");
else
document.write("node 25 not found" + "</br>");
level = getLevel(root, 27);
if (level != 0)
document.write("level of node 27 is " + level + "</br>");
else
document.write("node 27 not found" + "</br>");
// This code is contributed by suresh07.
</script>
输出:
level of node 30 is 2
level of node 12 is 4
level of node 25 is 3
node 27 not found
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