给定一棵二叉树，任务是找到二叉树中偶数节点和奇数节点之间的绝对差。
例子：
Input:
5
/ \
2 6
/ \ \
1 4 8
/ / \
3 7 9
Output: 5
Explanation:
Sum of the odd value nodes is:
5 + 1 + 3 + 7 + 9 = 25
Sum of the even value nodes is:
2 + 6 + 4 + 8 = 20
Absolute difference = (25 – 20) = 5.
Input:
4
/ \
1 4
/ \ \
7 2 6
Output: 8
方法：
请按照以下步骤解决问题：
- 遍历树中的每个节点并检查该节点处的值是奇数还是偶数。
- 在访问每个节点后相应地更新oddSum和evenSum 。
- 完成树的遍历后，打印oddSum和evenSum之间的绝对差。
下面是上述方法的实现：
C++
// C++ implementation of
// the above approach
#include
using namespace std;
int oddsum = 0;
int evensum = 0;
int ans = 0;
struct node {
int data;
struct node* left;
struct node* right;
};
struct node* newnode(int data)
{
node* temp = new node();
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// Function calculate the sum of
// odd and even value node
void OddEvenDifference(struct node*
root)
{
// If root is NULL
if (root == NULL) {
return;
}
else {
// Check if current root
// is odd or even
if (root->data % 2 == 0) {
evensum += root->data;
}
else {
oddsum += root->data;
}
// Call on the left subtree
OddEvenDifference(root->left);
// Call on the right subtree
OddEvenDifference(root->right);
}
}
// Driver Code
int main()
{
node* root = newnode(5);
root->left = newnode(2);
root->right = newnode(6);
root->left->left = newnode(1);
root->left->right = newnode(4);
root->left->right->left
= newnode(3);
root->right->right = newnode(8);
root->right->right->right
= newnode(9);
root->right->right->left
= newnode(7);
OddEvenDifference(root);
cout << abs(oddsum - evensum)
<< endl;
} Java
// Java implementation of
// the above approach
class GFG{
static int oddsum = 0;
static int evensum = 0;
static int ans = 0;
static class node
{
int data;
node left;
node right;
};
static node newnode(int data)
{
node temp = new node();
temp.data = data;
temp.left = temp.right = null;
return temp;
}
// Function calculate the sum of
// odd and even value node
static void OddEvenDifference(node root)
{
// If root is null
if (root == null)
{
return;
}
else
{
// Check if current root
// is odd or even
if (root.data % 2 == 0)
{
evensum += root.data;
}
else
{
oddsum += root.data;
}
// Call on the left subtree
OddEvenDifference(root.left);
// Call on the right subtree
OddEvenDifference(root.right);
}
}
// Driver Code
public static void main(String[] args)
{
node root = newnode(5);
root.left = newnode(2);
root.right = newnode(6);
root.left.left = newnode(1);
root.left.right = newnode(4);
root.left.right.left = newnode(3);
root.right.right = newnode(8);
root.right.right.right = newnode(9);
root.right.right.left = newnode(7);
OddEvenDifference(root);
System.out.print(Math.abs(
oddsum - evensum) + "\n");
}
}
// This code is contributed by amal kumar choubeyPython3
# Python3 implementation of
# the above approach
oddsum = 0
evensum = 0
class Node:
def __init__(self, data):
self.left = None
self.right = None
self.data = data
def newnode(data):
temp = Node(data)
return temp
# Function calculate the sum of
# odd and even value node
def OddEvenDifference(root):
global evensum, oddsum
# If root is NULL
if (root == None):
return
else:
# Check if current root
# is odd or even
if (root.data % 2 == 0):
evensum += root.data
else:
oddsum += root.data
# Call on the left subtree
OddEvenDifference(root.left)
# Call on the right subtree
OddEvenDifference(root.right)
# Driver code
if __name__=="__main__":
root = newnode(5)
root.left = newnode(2)
root.right = newnode(6)
root.left.left = newnode(1)
root.left.right = newnode(4)
root.left.right.left= newnode(3)
root.right.right = newnode(8)
root.right.right.right= newnode(9)
root.right.right.left= newnode(7)
OddEvenDifference(root)
print(abs(oddsum - evensum))
# This code is contributed by rutvik_56C#
// C# implementation of
// the above approach
using System;
class GFG{
static int oddsum = 0;
static int evensum = 0;
//static int ans = 0;
class node
{
public int data;
public node left;
public node right;
};
static node newnode(int data)
{
node temp = new node();
temp.data = data;
temp.left = temp.right = null;
return temp;
}
// Function calculate the sum of
// odd and even value node
static void OddEvenDifference(node root)
{
// If root is null
if (root == null)
{
return;
}
else
{
// Check if current root
// is odd or even
if (root.data % 2 == 0)
{
evensum += root.data;
}
else
{
oddsum += root.data;
}
// Call on the left subtree
OddEvenDifference(root.left);
// Call on the right subtree
OddEvenDifference(root.right);
}
}
// Driver Code
public static void Main(String[] args)
{
node root = newnode(5);
root.left = newnode(2);
root.right = newnode(6);
root.left.left = newnode(1);
root.left.right = newnode(4);
root.left.right.left = newnode(3);
root.right.right = newnode(8);
root.right.right.right = newnode(9);
root.right.right.left = newnode(7);
OddEvenDifference(root);
Console.Write(Math.Abs(
oddsum - evensum) + "\n");
}
}
// This code is contributed by amal kumar choubey输出：
5
时间复杂度： O(N)
辅助空间： O(1)
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