二叉树中给定节点除相邻节点外的总和
给定一个BT和一个关键节点，求BT中的总和，除了那些与关键节点相邻的节点。
例子：
1. 使用前序遍历树。
2.如果当前节点与键相邻，则不要将其添加到最终总和中。
3. 如果当前节点是键，则不要将其子节点添加到最终总和中。
4. 如果键不存在，则返回所有节点的总和。
C++
// C++ program to find total sum except a given Node in BT
#include
using namespace std;
struct Node {
int data;
struct Node *left, *right;
};
// insertion of Node in Tree
Node* getNode(int n)
{
struct Node* root = new Node;
root->data = n;
root->left = NULL;
root->right = NULL;
return root;
}
// sum of all element except those which are adjacent to key Node
void find_sum(Node* root, int key, int& sum, bool incl)
{
if (root) {
if (incl) {
sum += root->data;
if (root->left && root->left->data == key) {
sum -= root->data;
}
else if (root->right && root->right->data == key) {
sum -= root->data;
}
}
incl = root->data == key ? false : true;
find_sum(root->left, key, sum, incl);
find_sum(root->right, key, sum, incl);
}
}
// Driver code
int main()
{
struct Node* root = getNode(15);
root->left = getNode(13);
root->left->left = getNode(12);
root->left->left->left = getNode(11);
root->left->right = getNode(14);
root->right = getNode(20);
root->right->left = getNode(18);
root->right->right = getNode(24);
root->right->right->left = getNode(23);
root->right->right->right = getNode(25);
int key = 20;
int sum = 0;
find_sum(root, key, sum, true);
printf("%d ", sum);
return 0;
} Java
// Java program to find total sum except a given Node in BT
import java.util.*;
class Node
{
int data;
Node left, right;
// insertion of Node in Tree
Node(int key)
{
data = key;
left = right = null;
}
}
class GFG
{
static class cal
{
int sum = 0;
}
// sum of all element except those which are adjacent to key Node
public static void find_sum(Node root, int key, cal r, boolean incl)
{
if(root != null)
{
if(incl == true)
{
r.sum += root.data;
if((root.left != null) && (root.left.data == key))
{
r.sum -= root.data;
}
else
if((root.right != null) && (root.right.data == key))
{
r.sum -= root.data;
}
}
if(root.data == key)
incl = false;
else
incl = true;
find_sum(root.left, key, r, incl);
find_sum(root.right, key, r, incl);
}
}
// Driver code
public static void main (String[] args)
{
Node root = new Node(15);
root.left = new Node(13);
root.left.left = new Node(12);
root.left.left.left = new Node(11);
root.left.right = new Node(14);
root.right = new Node(20);
root.right.left = new Node(18);
root.right.right = new Node(24);
root.right.right.right = new Node(25);
root.right.right.left = new Node(23);
int key = 20;
cal obj = new cal();
find_sum(root, key, obj, true);
System.out.print(obj.sum);
}
}Python
# Python program to find total
# sum except a given Node in BT
class Node:
def __init__(self, data):
self.left = None
self.right = None
self.data = data
# Insertion of Node in Tree
def getNode(n):
root = Node(n)
return root
# sum of all element except
# those which are adjacent
# to key Node
def find_sum(root, key,
sum, incl):
if (root):
if(incl):
sum += root.data;
if (root.left and
root.left.data == key):
sum -= root.data;
elif (root.right and
root.right.data == key):
sum -= root.data;
if root.data == key:
incl = False
else:
incl = True
sum = find_sum(root.left, key,
sum, incl);
sum = find_sum(root.right, key,
sum, incl);
return sum
# Driver code
if __name__ == "__main__":
root = getNode(15);
root.left = getNode(13);
root.left.left = getNode(12);
root.left.left.left = getNode(11);
root.left.right = getNode(14);
root.right = getNode(20);
root.right.left = getNode(18);
root.right.right = getNode(24);
root.right.right.left = getNode(23);
root.right.right.right = getNode(25);
key = 20;
sum = 0;
sum = find_sum(root, key,
sum, True);
print(sum)
# This code is contributed by Rutvik_56C#
// C# program to find total sum
// except a given Node in BT
using System;
public class Node
{
public int data;
public Node left, right;
// insertion of Node in Tree
public Node(int key)
{
data = key;
left = right = null;
}
}
public class GFG
{
public class cal
{
public int sum = 0;
}
// sum of all element except those
// which are adjacent to key Node
public static void find_sum(Node root,
int key, cal r, bool incl)
{
if(root != null)
{
if(incl == true)
{
r.sum += root.data;
if((root.left != null) &&
(root.left.data == key))
{
r.sum -= root.data;
}
else
if((root.right != null) &&
(root.right.data == key))
{
r.sum -= root.data;
}
}
if(root.data == key)
incl = false;
else
incl = true;
find_sum(root.left, key, r, incl);
find_sum(root.right, key, r, incl);
}
}
// Driver code
public static void Main (String[] args)
{
Node root = new Node(15);
root.left = new Node(13);
root.left.left = new Node(12);
root.left.left.left = new Node(11);
root.left.right = new Node(14);
root.right = new Node(20);
root.right.left = new Node(18);
root.right.right = new Node(24);
root.right.right.right = new Node(25);
root.right.right.left = new Node(23);
int key = 20;
cal obj = new cal();
find_sum(root, key, obj, true);
Console.Write(obj.sum);
}
}
// This code is contributed Rajput-JiJavascript
输出：
118时间复杂度： O(n)，其中 n 是 BT 中的节点数。