所有节点到给定节点的距离总和

给定一棵二叉树和一个整数目标，表示一个节点的值，任务是找到所有节点到给定节点的距离之和。

例子：

Input: target = 3

Output: 19
Explanation:

Distance of Nodes 1, 6, 7 from the Node 3 = 1
Distance of the Node 2 from the Node 3 = 2
Distance of the Nodes 4, 5 from the Node 3 = 3
Distance of the Nodes 8, 9 from the Node 3 = 4
Sum of the distances = (1 + 1 + 1) + (2) + (3 + 3) + (4 + 4) = 19.

Input: target = 4

Output: 18

天真的方法：解决此问题的最简单方法是，每当在一个节点的左侧或右侧遍历一个节点时，则其子树的节点距离将减少1，其余节点与该节点的距离将减少增加1。
因此，以下关系给出了所有节点到一个节点的距离之和，即u ：

sumDists(u)= sumDists(parent(u)) – (Nodes in the left and right substree of u) + (N – Nodes in the left and right substree of u)

where, 
sumDists(u): Sum of distances of all nodes from the node u
sumDists(parent(u)): Sum of distances of all nodes from the parent node of u

请按照以下步骤解决问题：

• 创建一个函数以查找给定节点(包括给定节点)的左右子树中的节点数。
• 创建一个函数以查找节点的深度之和，变量sum表示所有节点距目标的距离之和。
• 使用DFS(深度优先搜索)遍历树，并对每个节点执行以下操作：
  • 如果目标与当前节点匹配，则将sum更新为distance 。
  • 别的：
    • 如果root-> left不为null ，请找到左子树中的节点数，并将所有节点到root-> left节点的距离之和作为tempSum传递。
    • 如果root-> right不为null ，则找到右子树中的节点数，并将所有节点到root-> rightnode的距离之和作为tempSum传递。
• 到达目标节点后，打印节点到目标节点的距离之和。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Structure of a
// Binary Tree Node
class TreeNode {
public:
    int data;
    TreeNode* left;
    TreeNode* right;
};
 
// Function that allocates a new node
// with the given data and NULL to its
// left and right pointers
TreeNode* newNode(int data)
{
    // Allocate the node
    TreeNode* Node = new TreeNode();
 
    // Allocate Memory
    Node->data = data;
    Node->left = NULL;
    Node->right = NULL;
 
    return (Node);
}
 
// Function which calculates sum
// of depths of all nodes
int sumofdepth(TreeNode* root, int l)
{
    // Base Case
    if (root == NULL)
        return 0;
 
    // Return recurssively
    return l + sumofdepth(root->left,
                          l + 1)
           + sumofdepth(root->right,
                        l + 1);
}
 
// Function to count of nodes
// in the left and right subtree
int Noofnodes(TreeNode* root)
{
    // Base Case
    if (root == NULL)
        return 0;
 
    // Return recurssively
    return Noofnodes(root->left)
           + Noofnodes(root->right)
           + 1;
}
 
// Stores the sum of distances
// of all nodes from given node
int sum = 0;
 
// Function to find sum of distances
// of all nodes from a given node
void distance(TreeNode* root,
              int target,
              int distancesum,
              int n)
{
    // If target node matches
    // with the current node
    if (root->data == target) {
        sum = distancesum;
        return;
    }
 
    // If left of current node exists
    if (root->left) {
 
        // Count number of nodes
        // in the left subtree
        int nodes = Noofnodes(
            root->left);
 
        // Update sum
        int tempsum = distancesum
                      - nodes
                      + (n - nodes);
 
        // Recur for the left subtree
        distance(root->left, target,
                 tempsum, n);
    }
 
    // If right is not null
    if (root->right) {
 
        // Find number of nodes
        // in the left subtree
        int nodes = Noofnodes(
            root->right);
 
        // Applying the formula given
        // in the approach
        int tempsum = distancesum
                      - nodes + (n - nodes);
 
        // Recur for the right subtree
        distance(root->right, target,
                 tempsum, n);
    }
}
 
// Driver Code
int main()
{
    // Input tree
    TreeNode* 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(6);
    root->right->right = newNode(7);
    root->left->left->left = newNode(8);
    root->left->left->right = newNode(9);
 
    int target = 3;
 
    // Sum of depth of all
    // nodes from root node
    int distanceroot
        = sumofdepth(root, 0);
 
    // Number of nodes in the
    // left and right subtree
    int totalnodes = Noofnodes(root);
 
    distance(root, target, distanceroot,
             totalnodes);
 
    // Print the sum of distances
    cout << sum;
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Structure of a
// Binary Tree Node
static class TreeNode
{
    int data;
    TreeNode left, right;
}
 
// Function that allocates a new node
// with the given data and NULL to its
// left and right pointers
static TreeNode newNode(int data)
{
    TreeNode Node = new TreeNode();
    Node.data = data;
    Node.left = Node.right = null;
    return (Node);
}
 
// Function which calculates sum
// of depths of all nodes
static int sumofdepth(TreeNode root, int l)
{
     
    // Base Case
    if (root == null)
        return 0;
         
    // Return recurssively
    return l + sumofdepth(root.left, l + 1) +
              sumofdepth(root.right, l + 1);
}
 
// Function to count of nodes
// in the left and right subtree
static int Noofnodes(TreeNode root)
{
     
    // Base Case
    if (root == null)
        return 0;
 
    // Return recurssively
    return Noofnodes(root.left) +
          Noofnodes(root.right) + 1;
}
 
// Stores the sum of distances
// of all nodes from given node
public static int sum = 0;
 
// Function to find sum of distances
// of all nodes from a given node
static void distance(TreeNode root, int target,
                     int distancesum, int n)
{
     
    // If target node matches
    // with the current node
    if (root.data == target)
    {
        sum = distancesum;
        return;
    }
 
    // If left of current node exists
    if (root.left != null)
    {
         
        // Count number of nodes
        // in the left subtree
        int nodes = Noofnodes(root.left);
 
        // Update sum
        int tempsum = distancesum - nodes +
                               (n - nodes);
 
        // Recur for the left subtree
        distance(root.left, target, tempsum, n);
    }
 
    // If right is not null
    if (root.right != null)
    {
         
        // Find number of nodes
        // in the left subtree
        int nodes = Noofnodes(root.right);
 
        // Applying the formula given
        // in the approach
        int tempsum = distancesum - nodes +
                               (n - nodes);
 
        // Recur for the right subtree
        distance(root.right, target, tempsum, n);
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Input tree
    TreeNode 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(6);
    root.right.right = newNode(7);
    root.left.left.left = newNode(8);
    root.left.left.right = newNode(9);
 
    int target = 3;
 
    // Sum of depth of all
    // nodes from root node
    int distanceroot = sumofdepth(root, 0);
 
    // Number of nodes in the
    // left and right subtree
    int totalnodes = Noofnodes(root);
 
    distance(root, target, distanceroot,
             totalnodes);
 
    // Print the sum of distances
    System.out.println(sum);
}
}
 
// This code is contributed by Dharanendra L V

Python3

# Python3 program for the above approach
 
# Structure of a
# Binary Tree Node
class TreeNode:
    def __init__(self, x):
        self.data = x
        self.left = None
        self.right = None
 
# Function which calculates sum
# of depths of all nodes
def sumofdepth(root, l):
   
    # Base Case
    if (root == None):
        return 0
 
    # Return recurssively
    return l + sumofdepth(root.left, l + 1)+ sumofdepth(root.right, l + 1)
 
# Function to count of nodes
# in the left and right subtree
def Noofnodes(root):
   
    # Base Case
    if (root == None):
        return 0
 
    # Return recurssively
    return Noofnodes(root.left) + Noofnodes(root.right) + 1
 
# Stores the sum of distances
# of all nodes from given node
sum = 0
 
# Function to find sum of distances
# of all nodes from a given node
def distance(root, target, distancesum, n):
    global sum
     
    # If target node matches
    # with the current node
    if (root.data == target):
        sum = distancesum
        return
 
    # If left of current node exists
    if (root.left):
 
        # Count number of nodes
        # in the left subtree
        nodes = Noofnodes(root.left)
 
        # Update sum
        tempsum = distancesum - nodes + (n - nodes)
 
        # Recur for the left subtree
        distance(root.left, target, tempsum, n)
 
    # If right is not null
    if (root.right):
 
        # Find number of nodes
        # in the left subtree
        nodes = Noofnodes(root.right)
 
        # Applying the formula given
        # in the approach
        tempsum = distancesum - nodes + (n - nodes)
 
        # Recur for the right subtree
        distance(root.right, target, tempsum, n)
 
# Driver Code
if __name__ == '__main__':
   
    # Input tree
    root = TreeNode(1)
    root.left = TreeNode(2)
    root.right = TreeNode(3)
    root.left.left = TreeNode(4)
    root.left.right = TreeNode(5)
    root.right.left = TreeNode(6)
    root.right.right = TreeNode(7)
    root.left.left.left = TreeNode(8)
    root.left.left.right = TreeNode(9)
    target = 3
 
    # Sum of depth of all
    # nodes from root node
    distanceroot = sumofdepth(root, 0)
 
    # Number of nodes in the
    # left and right subtree
    totalnodes = Noofnodes(root)
    distance(root, target, distanceroot, totalnodes)
 
    # Prthe sum of distances
    print (sum)
 
    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
 
public class GFG{
 
// Structure of a
// Binary Tree Node
class TreeNode
{
    public int data;
    public TreeNode left, right;
}
 
// Function that allocates a new node
// with the given data and NULL to its
// left and right pointers
static TreeNode newNode(int data)
{
    TreeNode Node = new TreeNode();
    Node.data = data;
    Node.left = Node.right = null;
    return (Node);
}
 
// Function which calculates sum
// of depths of all nodes
static int sumofdepth(TreeNode root, int l)
{
     
    // Base Case
    if (root == null)
        return 0;
         
    // Return recurssively
    return l + sumofdepth(root.left, l + 1) +
              sumofdepth(root.right, l + 1);
}
 
// Function to count of nodes
// in the left and right subtree
static int Noofnodes(TreeNode root)
{
     
    // Base Case
    if (root == null)
        return 0;
 
    // Return recurssively
    return Noofnodes(root.left) +
          Noofnodes(root.right) + 1;
}
 
// Stores the sum of distances
// of all nodes from given node
public static int sum = 0;
 
// Function to find sum of distances
// of all nodes from a given node
static void distance(TreeNode root, int target,
                     int distancesum, int n)
{
     
    // If target node matches
    // with the current node
    if (root.data == target)
    {
        sum = distancesum;
        return;
    }
 
    // If left of current node exists
    if (root.left != null)
    {
         
        // Count number of nodes
        // in the left subtree
        int nodes = Noofnodes(root.left);
 
        // Update sum
        int tempsum = distancesum - nodes +
                               (n - nodes);
 
        // Recur for the left subtree
        distance(root.left, target, tempsum, n);
    }
 
    // If right is not null
    if (root.right != null)
    {
         
        // Find number of nodes
        // in the left subtree
        int nodes = Noofnodes(root.right);
 
        // Applying the formula given
        // in the approach
        int tempsum = distancesum - nodes +
                               (n - nodes);
 
        // Recur for the right subtree
        distance(root.right, target, tempsum, n);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Input tree
    TreeNode 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(6);
    root.right.right = newNode(7);
    root.left.left.left = newNode(8);
    root.left.left.right = newNode(9);
 
    int target = 3;
 
    // Sum of depth of all
    // nodes from root node
    int distanceroot = sumofdepth(root, 0);
 
    // Number of nodes in the
    // left and right subtree
    int totalnodes = Noofnodes(root);
 
    distance(root, target, distanceroot,
             totalnodes);
 
    // Print the sum of distances
    Console.WriteLine(sum);
}
}
 
// This code is contributed by shikhasingrajput

输出：

19

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：可以通过添加一个额外的变量(例如size)来优化上述方法，以表示节点结构中其左侧和右侧子树中的节点数。这样可以将计算子树大小的任务减少到恒定的计算时间

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Structure of a
// binary tree node
class TreeNode {
public:
    int data, size;
    TreeNode* left;
    TreeNode* right;
};
 
// Function that allocates a new node
// with the given data and NULL to
// its left and right pointers
TreeNode* newNode(int data)
{
    TreeNode* Node = new TreeNode();
    Node->data = data;
    Node->left = NULL;
    Node->right = NULL;
 
    // Return newly created node
    return (Node);
}
 
// Function to count the number of
// nodes in the left and right subtrees
pair sumofsubtree(TreeNode* root)
{
    // Initialize a pair that stores
    // the pair {number of nodes, depth}
    pair p = make_pair(1, 0);
 
    // Finding the number of nodes
    // in the left subtree
    if (root->left) {
        pair ptemp
            = sumofsubtree(root->left);
 
        p.second += ptemp.first
                    + ptemp.second;
        p.first += ptemp.first;
    }
 
    // Find the number of nodes
    // in the right subtree
    if (root->right) {
 
        pair ptemp
            = sumofsubtree(root->right);
 
        p.second += ptemp.first
                    + ptemp.second;
        p.first += ptemp.first;
    }
 
    // Filling up size field
    root->size = p.first;
    return p;
}
 
// Stores the sum of distances of all
// nodes from the given node
int sum = 0;
 
// Function to find the total distance
void distance(TreeNode* root, int target,
              int distancesum, int n)
{
    // If target node matches with
    // the current node
    if (root->data == target) {
        sum = distancesum;
    }
 
    // If root->left is not null
    if (root->left) {
 
        // Update sum
        int tempsum = distancesum
                      - root->left->size
                      + (n - root->left->size);
 
        // Recur for the left subtree
        distance(root->left, target,
                 tempsum, n);
    }
 
    // If root->right is not null
    if (root->right) {
 
        // Apply the formula given
        // in the approach
        int tempsum = distancesum
                      - root->right->size
                      + (n - root->right->size);
 
        // Recur for the right subtree
        distance(root->right, target,
                 tempsum, n);
    }
}
 
// Driver Code
int main()
{
    // Input tree
    TreeNode* 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(6);
    root->right->right = newNode(7);
    root->left->left->left = newNode(8);
    root->left->left->right = newNode(9);
 
    int target = 3;
 
    pair p = sumofsubtree(root);
 
    // Total number of nodes
    int totalnodes = p.first;
 
    distance(root, target, p.second,
             totalnodes);
 
    // Print the sum of distances
    cout << sum << endl;
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
    static class pair
    {
        int first, second;
        public pair(int first, int second) 
        {
            this.first = first;
            this.second = second;
        }   
    }
   
// Structure of a
// binary tree node
static class TreeNode
{
    int data, size;
    TreeNode left;
    TreeNode right;
};
 
// Function that allocates a new node
// with the given data and null to
// its left and right pointers
static TreeNode newNode(int data)
{
    TreeNode Node = new TreeNode();
    Node.data = data;
    Node.left = null;
    Node.right = null;
 
    // Return newly created node
    return (Node);
}
 
// Function to count the number of
// nodes in the left and right subtrees
static pair sumofsubtree(TreeNode root)
{
   
    // Initialize a pair that stores
    // the pair {number of nodes, depth}
    pair p = new pair(1, 0);
 
    // Finding the number of nodes
    // in the left subtree
    if (root.left != null)
    {
        pair ptemp
            = sumofsubtree(root.left);
 
        p.second += ptemp.first
                    + ptemp.second;
        p.first += ptemp.first;
    }
 
    // Find the number of nodes
    // in the right subtree
    if (root.right != null)
    {
        pair ptemp
            = sumofsubtree(root.right);
 
        p.second += ptemp.first
                    + ptemp.second;
        p.first += ptemp.first;
    }
 
    // Filling up size field
    root.size = p.first;
    return p;
}
 
// Stores the sum of distances of all
// nodes from the given node
static int sum = 0;
 
// Function to find the total distance
static void distance(TreeNode root, int target,
              int distancesum, int n)
{
   
    // If target node matches with
    // the current node
    if (root.data == target)
    {
        sum = distancesum;
    }
 
    // If root.left is not null
    if (root.left != null)
    {
 
        // Update sum
        int tempsum = distancesum
                      - root.left.size
                      + (n - root.left.size);
 
        // Recur for the left subtree
        distance(root.left, target,
                 tempsum, n);
    }
 
    // If root.right is not null
    if (root.right != null)
    {
 
        // Apply the formula given
        // in the approach
        int tempsum = distancesum
                      - root.right.size
                      + (n - root.right.size);
 
        // Recur for the right subtree
        distance(root.right, target,
                 tempsum, n);
    }
}
 
// Driver Code
public static void main(String[] args)
{
   
    // Input tree
    TreeNode 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(6);
    root.right.right = newNode(7);
    root.left.left.left = newNode(8);
    root.left.left.right = newNode(9);
    int target = 3;
    pair p = sumofsubtree(root);
 
    // Total number of nodes
    int totalnodes = p.first;
    distance(root, target, p.second,
             totalnodes);
 
    // Print the sum of distances
    System.out.print(sum +"\n");
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for the above approach
 
# Stores the sum of distances of all
# nodes from the given node
sum = 0
 
# Structure of a
# binary tree node
class TreeNode:
     
    def __init__(self, data):
         
        self.data = data
        self.size = 0
        self.left = None
        self.right = None
 
# Function to count the number of
# nodes in the left and right subtrees
def sumofsubtree(root):
     
    # Initialize a pair that stores
    # the pair {number of nodes, depth}
    p =  [1, 0]
 
    # Finding the number of nodes
    # in the left subtree
    if (root.left):
        ptemp = sumofsubtree(root.left)
        p[1] += ptemp[0] + ptemp[1]
        p[0] += ptemp[0]
 
    # Find the number of nodes
    # in the right subtree
    if (root.right):
        ptemp = sumofsubtree(root.right)
        p[1] += ptemp[0] + ptemp[1]
        p[0] += ptemp[0]
 
    # Filling up size field
    root.size = p[0]
    return p
 
# Function to find the total distance
def distance(root, target, distancesum, n):
     
    global sum
     
    # If target node matches with
    # the current node
    if (root.data == target):
        sum = distancesum
 
    # If root.left is not null
    if (root.left):
         
        # Update sum
        tempsum = (distancesum - root.left.size +
                            (n - root.left.size))
 
        # Recur for the left subtree
        distance(root.left, target, tempsum, n)
 
    # If root.right is not null
    if (root.right):
         
        # Apply the formula given
        # in the approach
        tempsum = (distancesum - root.right.size +
                            (n - root.right.size))
 
        # Recur for the right subtree
        distance(root.right, target, tempsum, n)
 
# Driver Code
if __name__ == '__main__':
     
    # Input tree
    root = TreeNode(1)
    root.left = TreeNode(2)
    root.right = TreeNode(3)
    root.left.left = TreeNode(4)
    root.left.right = TreeNode(5)
    root.right.left = TreeNode(6)
    root.right.right = TreeNode(7)
    root.left.left.left = TreeNode(8)
    root.left.left.right = TreeNode(9)
 
    target = 3
 
    p = sumofsubtree(root)
 
    # Total number of nodes
    totalnodes = p[0]
 
    distance(root, target, p[1], totalnodes)
 
    # Print the sum of distances
    print(sum)
 
# This code is contributed by ipg2016107

C＃

// C# program for the above approach
using System;
public class GFG
{
    class pair
    {
        public int first, second;
        public pair(int first, int second) 
        {
            this.first = first;
            this.second = second;
        }   
    }
   
// Structure of a
// binary tree node
class TreeNode
{
    public int data, size;
    public TreeNode left;
    public TreeNode right;
};
 
// Function that allocates a new node
// with the given data and null to
// its left and right pointers
static TreeNode newNode(int data)
{
    TreeNode Node = new TreeNode();
    Node.data = data;
    Node.left = null;
    Node.right = null;
 
    // Return newly created node
    return (Node);
}
 
// Function to count the number of
// nodes in the left and right subtrees
static pair sumofsubtree(TreeNode root)
{
   
    // Initialize a pair that stores
    // the pair {number of nodes, depth}
    pair p = new pair(1, 0);
 
    // Finding the number of nodes
    // in the left subtree
    if (root.left != null)
    {
        pair ptemp
            = sumofsubtree(root.left);
 
        p.second += ptemp.first
                    + ptemp.second;
        p.first += ptemp.first;
    }
 
    // Find the number of nodes
    // in the right subtree
    if (root.right != null)
    {
        pair ptemp
            = sumofsubtree(root.right);
 
        p.second += ptemp.first
                    + ptemp.second;
        p.first += ptemp.first;
    }
 
    // Filling up size field
    root.size = p.first;
    return p;
}
 
// Stores the sum of distances of all
// nodes from the given node
static int sum = 0;
 
// Function to find the total distance
static void distance(TreeNode root, int target,
              int distancesum, int n)
{
   
    // If target node matches with
    // the current node
    if (root.data == target)
    {
        sum = distancesum;
    }
 
    // If root.left is not null
    if (root.left != null)
    {
 
        // Update sum
        int tempsum = distancesum
                      - root.left.size
                      + (n - root.left.size);
 
        // Recur for the left subtree
        distance(root.left, target,
                 tempsum, n);
    }
 
    // If root.right is not null
    if (root.right != null)
    {
 
        // Apply the formula given
        // in the approach
        int tempsum = distancesum
                      - root.right.size
                      + (n - root.right.size);
 
        // Recur for the right subtree
        distance(root.right, target,
                 tempsum, n);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
   
    // Input tree
    TreeNode 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(6);
    root.right.right = newNode(7);
    root.left.left.left = newNode(8);
    root.left.left.right = newNode(9);
    int target = 3;
    pair p = sumofsubtree(root);
 
    // Total number of nodes
    int totalnodes = p.first;
    distance(root, target, p.second,
             totalnodes);
 
    // Print the sum of distances
    Console.Write(sum +"\n");
}
}
 
// This code is contributed by shikhasingrajput

输出：

19

时间复杂度： O(N)
辅助空间： O(1)