给定一棵二叉树，任务是检查该二叉树是否为偶数奇数二叉树。
A Binary Tree is called an Even-Odd Tree when all the nodes which are at even levels have even values (assuming root to be at level 0) and all the nodes which are at odd levels have odd values.
例子：
Input:
Output: YES
Explanation:
Only node on level 0 (even) is 2 (even).
Nodes present in level 1 are 3 and 9 (both odd).
Nodes present in level 2 are 4, 10 and 6 (all even).
Therefore, the Binary tree is an odd-even binary tree.
Input:
Output: NO
方法：请按照以下步骤解决问题：
- 这个想法是执行级别顺序遍历，并检查偶数级别上存在的节点是否为偶数，奇数级别上存在的节点是否为奇数。
- 如果发现任何处于奇数级的节点具有奇数值，反之亦然，则打印“ NO ”。
- 否则，在完全遍历树后，打印“是”。
下面是上述方法的实现：
C++
2
/ \
3 9
/ \ \
4 10 6
Java
4
/ \
3 7
/ \ \
4 10 5
Python3
// C++ program for the above approach
#include
using namespace std;
struct Node
{
int val;
Node *left, *right;
};
struct Node* newNode(int data)
{
struct Node* temp = (struct Node*)malloc(sizeof(struct Node));
temp->val = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}
// Function to check if the
// tree is even-odd tree
bool isEvenOddBinaryTree(Node *root)
{
if (root == NULL)
return true;
// Stores nodes of each level
queue q;
q.push(root);
// Store the current level
// of the binary tree
int level = 0;
// Traverse until the
// queue is empty
while (!q.empty())
{
// Stores the number of nodes
// present in the current level
int size = q.size();
for(int i = 0; i < size; i++)
{
Node *node = q.front();
// Check if the level
// is even or odd
if (level % 2 == 0)
{
if (node->val % 2 == 1)
return false;
}
else if (level % 2 == 1)
{
if (node->val % 2 == 0)
return true;
}
// Add the nodes of the next
// level into the queue
if (node->left != NULL)
{
q.push(node->left);
}
if (node->right != NULL)
{
q.push(node->right);
}
}
// Increment the level count
level++;
}
return true;
}
// Driver Code
int main()
{
// Construct a Binary Tree
Node *root = NULL;
root = newNode(2);
root->left = newNode(3);
root->right = newNode(9);
root->left->left = newNode(4);
root->left->right = newNode(10);
root->right->right = newNode(6);
// Check if the binary tree
// is even-odd tree or not
if (isEvenOddBinaryTree(root))
cout << "YES";
else
cout << "NO";
}
// This code is contributed by ipg2016107 C#
// Java Program for the above approach
import java.util.*;
class GfG {
// Tree node
static class Node {
int val;
Node left, right;
}
// Function to return new tree node
static Node newNode(int data)
{
Node temp = new Node();
temp.val = data;
temp.left = null;
temp.right = null;
return temp;
}
// Function to check if the
// tree is even-odd tree
public static boolean
isEvenOddBinaryTree(Node root)
{
if (root == null)
return true;
// Stores nodes of each level
Queue q
= new LinkedList<>();
q.add(root);
// Store the current level
// of the binary tree
int level = 0;
// Traverse until the
// queue is empty
while (!q.isEmpty()) {
// Stores the number of nodes
// present in the current level
int size = q.size();
for (int i = 0; i < size; i++) {
Node node = q.poll();
// Check if the level
// is even or odd
if (level % 2 == 0) {
if (node.val % 2 == 1)
return false;
}
else if (level % 2 == 1) {
if (node.val % 2 == 0)
return false;
}
// Add the nodes of the next
// level into the queue
if (node.left != null) {
q.add(node.left);
}
if (node.right != null) {
q.add(node.right);
}
}
// Increment the level count
level++;
}
return true;
}
// Driver Code
public static void main(String[] args)
{
// Construct a Binary Tree
Node root = null;
root = newNode(2);
root.left = newNode(3);
root.right = newNode(9);
root.left.left = newNode(4);
root.left.right = newNode(10);
root.right.right = newNode(6);
// Check if the binary tree
// is even-odd tree or not
if (isEvenOddBinaryTree(root)) {
System.out.println("YES");
}
else {
System.out.println("NO");
}
}
} 输出：
# Python3 program for the above approach
# Tree node
class Node:
def __init__(self, data):
self.left = None
self.right = None
self.val = data
# Function to return new tree node
def newNode(data):
temp = Node(data)
return temp
# Function to check if the
# tree is even-odd tree
def isEvenOddBinaryTree(root):
if (root == None):
return True
q = []
# Stores nodes of each level
q.append(root)
# Store the current level
# of the binary tree
level = 0
# Traverse until the
# queue is empty
while (len(q) != 0):
# Stores the number of nodes
# present in the current level
size = len(q)
for i in range(size):
node = q[0]
q.pop(0)
# Check if the level
# is even or odd
if (level % 2 == 0):
if (node.val % 2 == 1):
return False
elif (level % 2 == 1):
if (node.val % 2 == 0):
return False
# Add the nodes of the next
# level into the queue
if (node.left != None):
q.append(node.left)
if (node.right != None):
q.append(node.right)
# Increment the level count
level += 1
return True
# Driver code
if __name__=="__main__":
# Construct a Binary Tree
root = None
root = newNode(2)
root.left = newNode(3)
root.right = newNode(9)
root.left.left = newNode(4)
root.left.right = newNode(10)
root.right.right = newNode(6)
# Check if the binary tree
# is even-odd tree or not
if (isEvenOddBinaryTree(root)):
print("YES")
else:
print("NO")
# This code is contributed by rutvik_56时间复杂度： O(N)
辅助空间： O(N)