给定二叉搜索树。任务是打印二进制搜索树的所有奇数节点。
例子：
Input :
5
/ \
3 7
/ \ / \
2 4 6 8
Output : 3 5 7
Input :
14
/ \
12 17
/ \ / \
8 13 16 19
Output : 13 17 19
方法：使用任何树遍历遍历二叉搜索树，并检查当前节点的值是否为奇数。如果是，则打印它，否则跳过该节点。
下面是上述方法的实现：
C++
// C++ program to print all odd node of BST
#include
using namespace std;
// create Tree
struct Node {
int key;
struct Node *left, *right;
};
// A utility function to create a new BST node
Node* newNode(int item)
{
Node* temp = new Node;
temp->key = item;
temp->left = temp->right = NULL;
return temp;
}
// A utility function to do inorder traversal of BST
void inorder(Node* root)
{
if (root != NULL) {
inorder(root->left);
printf("%d ", root->key);
inorder(root->right);
}
}
/* A utility function to insert a new node
with given key in BST */
Node* insert(Node* node, int key)
{
/* If the tree is empty, return a new node */
if (node == NULL)
return newNode(key);
/* Otherwise, recur down the tree */
if (key < node->key)
node->left = insert(node->left, key);
else
node->right = insert(node->right, key);
/* return the (unchanged) node pointer */
return node;
}
// Function to print all odd nodes
void oddNode(Node* root)
{
if (root != NULL) {
oddNode(root->left);
// if node is odd then print it
if (root->key % 2 != 0)
printf("%d ", root->key);
oddNode(root->right);
}
}
// Driver Code
int main()
{
/* Let us create following BST
5
/ \
3 7
/ \ / \
2 4 6 8 */
Node* root = NULL;
root = insert(root, 5);
root = insert(root, 3);
root = insert(root, 2);
root = insert(root, 4);
root = insert(root, 7);
root = insert(root, 6);
root = insert(root, 8);
oddNode(root);
return 0;
} Java
// Java program to print all odd node of BST
class GfG {
// create Tree
static class Node {
int key;
Node left, right;
}
// A utility function to create a new BST node
static Node newNode(int item)
{
Node temp = new Node();
temp.key = item;
temp.left = null;
temp.right = null;
return temp;
}
// A utility function to do inorder traversal of BST
static void inorder(Node root)
{
if (root != null) {
inorder(root.left);
System.out.print(root.key + " ");
inorder(root.right);
}
}
/* A utility function to insert a new node
with given key in BST */
static Node insert(Node node, int key)
{
/* If the tree is empty, return a new node */
if (node == null)
return newNode(key);
/* Otherwise, recur down the tree */
if (key < node.key)
node.left = insert(node.left, key);
else
node.right = insert(node.right, key);
/* return the (unchanged) node pointer */
return node;
}
// Function to print all odd nodes
static void oddNode(Node root)
{
if (root != null) {
oddNode(root.left);
// if node is odd then print it
if (root.key % 2 != 0)
System.out.print(root.key + " ");
oddNode(root.right);
}
}
// Driver Code
public static void main(String[] args)
{
/* Let us create following BST
5
/ \
3 7
/ \ / \
2 4 6 8 */
Node root = null;
root = insert(root, 5);
root = insert(root, 3);
root = insert(root, 2);
root = insert(root, 4);
root = insert(root, 7);
root = insert(root, 6);
root = insert(root, 8);
oddNode(root);
}
}Python3
# Python3 program to print all odd
# node of BST
# create Tree
# to create a new BST node
class newNode:
# Construct to create a new node
def __init__(self, key):
self.key = key
self.left = None
self.right = None
# A utility function to do inorder
# traversal of BST
def inorder( root) :
if (root != None):
inorder(root.left)
print(root.key, end = " ")
inorder(root.right)
""" A utility function to insert a
new node with given key in BST """
def insert(node, key):
""" If the tree is empty,
return a new node """
if (node == None):
return newNode(key)
""" Otherwise, recur down the tree """
if (key < node.key):
node.left = insert(node.left, key)
else:
node.right = insert(node.right, key)
""" return the (unchanged) node pointer """
return node
# Function to print all even nodes
def oddNode(root) :
if (root != None):
oddNode(root.left)
# if node is even then print it
if (root.key % 2 != 0):
print(root.key, end = " ")
oddNode(root.right)
# Driver Code
if __name__ == '__main__':
""" Let us create following BST
5
/ \
3 7
/ \ / \
2 4 6 8 """
root = None
root = insert(root, 5)
root = insert(root, 3)
root = insert(root, 2)
root = insert(root, 4)
root = insert(root, 7)
root = insert(root, 6)
root = insert(root, 8)
oddNode(root)
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)C#
// C# program to print all odd node of BST
using System;
public class GfG
{
// create Tree
class Node
{
public int key;
public Node left, right;
}
// A utility function to create a new BST node
static Node newNode(int item)
{
Node temp = new Node();
temp.key = item;
temp.left = null;
temp.right = null;
return temp;
}
// A utility function to do
// inorder traversal of BST
static void inorder(Node root)
{
if (root != null)
{
inorder(root.left);
Console.Write(root.key + " ");
inorder(root.right);
}
}
/* A utility function to insert a new node
with given key in BST */
static Node insert(Node node, int key)
{
/* If the tree is empty, return a new node */
if (node == null)
return newNode(key);
/* Otherwise, recur down the tree */
if (key < node.key)
node.left = insert(node.left, key);
else
node.right = insert(node.right, key);
/* return the (unchanged) node pointer */
return node;
}
// Function to print all odd nodes
static void oddNode(Node root)
{
if (root != null)
{
oddNode(root.left);
// if node is odd then print it
if (root.key % 2 != 0)
Console.Write(root.key + " ");
oddNode(root.right);
}
}
// Driver Code
public static void Main(String[] args)
{
/* Let us create following BST
5
/ \
3 7
/ \ / \
2 4 6 8 */
Node root = null;
root = insert(root, 5);
root = insert(root, 3);
root = insert(root, 2);
root = insert(root, 4);
root = insert(root, 7);
root = insert(root, 6);
root = insert(root, 8);
oddNode(root);
}
}
// This code has been contributed
// by PrinciRaj1992输出：
3 5 7
时间复杂度： O(n)，其中n为否。节点数
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