获取二叉树中节点的级别
给定一个二叉树和一个键，编写一个返回键级别的函数。
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，那么你的函数应该返回 0。
思路是从root开始，level为1。如果key与root的数据匹配，则返回level。否则递归调用级别为级别 + 1 的左右子树。
C++
// C++ program to Get Level of a
// node in a Binary Tree
#include
using namespace std;
/* A tree node structure */
struct node
{
int data;
struct node *left;
struct node *right;
};
/* Helper function for getLevel().
It returns level of the data if data is
present in tree, otherwise returns 0.*/
int getLevelUtil(struct node *node,
int data, int level)
{
if (node == NULL)
return 0;
if (node -> data == data)
return level;
int downlevel = getLevelUtil(node -> left,
data, level + 1);
if (downlevel != 0)
return downlevel;
downlevel = getLevelUtil(node->right,
data, level + 1);
return downlevel;
}
/* Returns level of given data value */
int getLevel(struct node *node, int data)
{
return getLevelUtil(node, data, 1);
}
/* Utility function to create a
new Binary Tree node */
struct node* newNode(int data)
{
struct node *temp = new struct node;
temp -> data = data;
temp -> left = NULL;
temp -> right = NULL;
return temp;
}
// Driver Code
int main()
{
struct node *root = new struct node;
int x;
/* Constructing tree given in
the above figure */
root = newNode(3);
root->left = newNode(2);
root->right = newNode(5);
root->left->left = newNode(1);
root->left->right = newNode(4);
for (x = 1; x <= 5; x++)
{
int level = getLevel(root, x);
if (level)
cout << "Level of "<< x << " is "
<< getLevel(root, x) << endl;
else
cout << x << "is not present in tree"
<< endl;
}
getchar();
return 0;
}
// This code is contributed
// by Akanksha Rai C
// C program to Get Level of a node in a Binary Tree
#include
#include
/* A tree node structure */
struct node
{
int data;
struct node *left;
struct node *right;
};
/* Helper function for getLevel().
It returns level of the data if data is
present in tree, otherwise returns 0.*/
int getLevelUtil(struct node *node,
int data, int level)
{
if (node == NULL)
return 0;
if (node->data == data)
return level;
int downlevel = getLevelUtil(node->left,
data, level+1);
if (downlevel != 0)
return downlevel;
downlevel = getLevelUtil(node->right,
data, level+1);
return downlevel;
}
/* Returns level of given data value */
int getLevel(struct node *node, int data)
{
return getLevelUtil(node,data,1);
}
/* Utility function to create
a new Binary Tree node */
struct node* newNode(int data)
{
struct node *temp = (struct node*)malloc(sizeof(struct node));
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}
/* Driver code */
int main()
{
struct node *root;
int x;
/* Constructing tree given
in the above figure */
root = newNode(3);
root->left = newNode(2);
root->right = newNode(5);
root->left->left = newNode(1);
root->left->right = newNode(4);
for (x = 1; x <=5; x++)
{
int level = getLevel(root, x);
if (level)
printf(" Level of %d is %d\n", x, getLevel(root, x));
else
printf(" %d is not present in tree \n", x);
}
getchar();
return 0;
} Java
// Java program to Get Level of a node in a Binary Tree
/* A tree node structure */
class Node {
int data;
Node left, right;
public Node(int d)
{
data = d;
left = right = null;
}
}
class BinaryTree {
Node root;
/* Helper function for getLevel().
It returns level of
the data if data is present in tree,
otherwise returns
0.*/
int getLevelUtil(Node node, int data, int level)
{
if (node == null)
return 0;
if (node.data == data)
return level;
int downlevel
= getLevelUtil(node.left, data, level + 1);
if (downlevel != 0)
return downlevel;
downlevel
= getLevelUtil(node.right, data, level + 1);
return downlevel;
}
/* Returns level of given data value */
int getLevel(Node node, int data)
{
return getLevelUtil(node, data, 1);
}
/* Driver code */
public static void main(String[] args)
{
BinaryTree tree = new BinaryTree();
/* Constructing tree given in the above figure */
tree.root = new Node(3);
tree.root.left = new Node(2);
tree.root.right = new Node(5);
tree.root.left.left = new Node(1);
tree.root.left.right = new Node(4);
for (int x = 1; x <= 5; x++)
{
int level = tree.getLevel(tree.root, x);
if (level != 0)
System.out.println(
"Level of " + x + " is "
+ tree.getLevel(tree.root, x));
else
System.out.println(
x + " is not present in tree");
}
}
}
// This code has been contributed by Mayank
// Jaiswal(mayank_24)Python3
# Python3 program to Get Level of a
# node in a Binary Tree
# Helper function that allocates a
# new node with the given data and
# None left and right pairs.
class newNode:
# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Helper function for getLevel(). It
# returns level of the data if data is
# present in tree, otherwise returns 0
def getLevelUtil(node, data, level):
if (node == None):
return 0
if (node.data == data):
return level
downlevel = getLevelUtil(node.left,
data, level + 1)
if (downlevel != 0):
return downlevel
downlevel = getLevelUtil(node.right,
data, level + 1)
return downlevel
# Returns level of given data value
def getLevel(node, data):
return getLevelUtil(node, data, 1)
# Driver Code
if __name__ == '__main__':
# Let us construct the Tree shown
# in the above figure
root = newNode(3)
root.left = newNode(2)
root.right = newNode(5)
root.left.left = newNode(1)
root.left.right = newNode(4)
for x in range(1, 6):
level = getLevel(root, x)
if (level):
print("Level of", x,
"is", getLevel(root, x))
else:
print(x, "is not present in tree")
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)C#
// C# program to Get Level of a node in a Binary Tree
using System;
/* A tree node structure */
public class Node {
public int data;
public Node left, right;
public Node(int d)
{
data = d;
left = right = null;
}
}
public class BinaryTree {
public Node root;
/* Helper function for getLevel().
It returns level of
the data if data is present in tree,
otherwise returns
0.*/
public virtual int getLevelUtil(Node node, int data,
int level)
{
if (node == null)
{
return 0;
}
if (node.data == data)
{
return level;
}
int downlevel
= getLevelUtil(node.left, data, level + 1);
if (downlevel != 0)
{
return downlevel;
}
downlevel
= getLevelUtil(node.right, data, level + 1);
return downlevel;
}
/* Returns level of given data value */
public virtual int getLevel(Node node, int data)
{
return getLevelUtil(node, data, 1);
}
/* Driver code */
public static void Main(string[] args)
{
BinaryTree tree = new BinaryTree();
/* Constructing tree given in the above figure */
tree.root = new Node(3);
tree.root.left = new Node(2);
tree.root.right = new Node(5);
tree.root.left.left = new Node(1);
tree.root.left.right = new Node(4);
for (int x = 1; x <= 5; x++)
{
int level = tree.getLevel(tree.root, x);
if (level != 0)
{
Console.WriteLine(
"Level of " + x + " is "
+ tree.getLevel(tree.root, x));
}
else
{
Console.WriteLine(
x + " is not present in tree");
}
}
}
}
// This code is contributed by Shrikant13Javascript
C++
/* C++ program to print level in which X is present
in binary tree using STL */
#include
using namespace std;
// A Binary Tree Node
struct Node {
int data;
struct Node *left, *right;
};
int printLevel(Node* root, int X)
{
Node* node;
// Base Case
if (root == NULL)
return 0;
// Create an empty queue for level order traversal
queue q;
// Create a var represent current level of tree
int currLevel = 1;
// Enqueue Root and initialize height
q.push(root);
while (q.empty() == false) {
// Print front of queue and remove it from queue
int size = q.size();
// traversing level by level
while (size--) {
node = q.front();
if (node->data == X) {
return currLevel;
}
q.pop();
/* Enqueue left child */
if (node->left != NULL)
q.push(node->left);
/*Enqueue right child */
if (node->right != NULL)
q.push(node->right);
}
// increment the level
currLevel++;
}
// if element not found
return 0;
}
// Utility function to create a new tree node
Node* newNode(int data)
{
Node* temp = new Node;
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// Driver program to test above functions
int main()
{
// Let us create binary tree shown in above diagram
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(7);
root->right->right = newNode(6);
cout< 输出
Level of 1 is 3
Level of 2 is 2
Level of 3 is 1
Level of 4 is 3
Level of 5 is 2getLevel() 的时间复杂度为 O(n)，其中 n 是给定二叉树中的节点数。
替代方法：给定问题可以借助给定二叉树的级别顺序遍历来解决。
C++
/* C++ program to print level in which X is present
in binary tree using STL */
#include
using namespace std;
// A Binary Tree Node
struct Node {
int data;
struct Node *left, *right;
};
int printLevel(Node* root, int X)
{
Node* node;
// Base Case
if (root == NULL)
return 0;
// Create an empty queue for level order traversal
queue q;
// Create a var represent current level of tree
int currLevel = 1;
// Enqueue Root and initialize height
q.push(root);
while (q.empty() == false) {
// Print front of queue and remove it from queue
int size = q.size();
// traversing level by level
while (size--) {
node = q.front();
if (node->data == X) {
return currLevel;
}
q.pop();
/* Enqueue left child */
if (node->left != NULL)
q.push(node->left);
/*Enqueue right child */
if (node->right != NULL)
q.push(node->right);
}
// increment the level
currLevel++;
}
// if element not found
return 0;
}
// Utility function to create a new tree node
Node* newNode(int data)
{
Node* temp = new Node;
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// Driver program to test above functions
int main()
{
// Let us create binary tree shown in above diagram
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(7);
root->right->right = newNode(6);
cout< 输出：3
时间复杂度：O(n)其中 n 是二叉树中的节点数
辅助空间：O(n) ，其中 n 是二叉树中的节点数