连接两个给定节点的路径中所有奇数节点的总和

给定一棵二叉树和该二叉树的两个节点。在连接两个给定节点的路径中找到所有具有奇数值的节点的总和。

例如：在上面的二叉树中，节点之间的路径中所有奇数节点的总和$5$和$6$将是5 + 1 + 3 = 9 。

来源：亚马逊校园面试体验
方法：想法是首先使用以下概念找到二叉树的两个给定节点之间的路径：打印任意两个节点之间的路径。
曾经，我们有两个给定节点之间的路径，计算该路径中所有奇值节点的总和并打印它。
下面是上述方法的实现：

C++

// C++ program to find sum of all odd nodes
// in the path connecting two given nodes
 
#include
using namespace std;
 
// Binary Tree node
struct Node
{
    int data;
    struct Node* left;
    struct Node* right;
};
 
// Utility function to create a
// new Binary Tree node
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = NULL;
    node->right = NULL;
     
    return node;
}
 
// Function to check if there is a path from root
// to the given node. It also populates
// 'arr' with the given path
bool getPath(Node* root, vector& arr, int x)
{
    // if root is NULL
    // there is no path
    if (!root)
        return false;
 
    // push the node's value in 'arr'
    arr.push_back(root->data);
 
    // if it is the required node
    // return true
    if (root->data == x)
        return true;
 
    // else check whether the required node lies
    // in the left subtree or right subtree of
    // the current node
    if (getPath(root->left, arr, x) || getPath(root->right, arr, x))
        return true;
 
    // required node does not lie either in the
    // left or right subtree of the current node
    // Thus, remove current node's value from
    // 'arr'and then return false
    arr.pop_back();
    return false;
}
 
// Function to get the sum of odd nodes in the
// path between any two nodes in a binary tree
int sumOddNodes(Node* root, int n1, int n2)
{
    // vector to store the path of
    // first node n1 from root
    vector path1;
 
    // vector to store the path of
    // second node n2 from root
    vector path2;
 
    getPath(root, path1, n1);
    getPath(root, path2, n2);
 
    int intersection = -1;
 
    // Get intersection point
    int i = 0, j = 0;
    while (i != path1.size() || j != path2.size()) {
 
        // Keep moving forward until no intersection
        // is found
        if (i == j && path1[i] == path2[j]) {
            i++;
            j++;
        }
        else {
            intersection = j - 1;
            break;
        }
    }
     
    int sum = 0;
     
    // calculate sum of ODD nodes from the path
    for (int i = path1.size() - 1; i > intersection; i--)
        if(path1[i]%2)
            sum += path1[i];
 
    for (int i = intersection; i < path2.size(); i++)
        if(path2[i]%2)
            sum += path2[i];
             
    return sum;       
}
 
// Driver Code
int main()
{
    struct Node* root = newNode(1);
     
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->right = newNode(6);
     
    int node1 = 5;
    int node2 = 6;
     
    cout<

---

Java

// Java program to find sum of all odd nodes
// in the path connecting two given nodes
 
import java.util.*;
class solution
{
 
// Binary Tree node
static class Node
{
    int data;
     Node left;
     Node right;
}
 
// Utility function to create a
// new Binary Tree node
 static Node newNode(int data)
{
     Node node = new Node();
    node.data = data;
    node.left = null;
    node.right = null;
     
    return node;
}
 
// Function to check if there is a path from root
// to the given node. It also populates
// 'arr' with the given path
static boolean getPath(Node root, Vector arr, int x)
{
    // if root is null
    // there is no path
    if (root==null)
        return false;
 
    // push the node's value in 'arr'
    arr.add(root.data);
 
    // if it is the required node
    // return true
    if (root.data == x)
        return true;
 
    // else check whether the required node lies
    // in the left subtree or right subtree of
    // the current node
    if (getPath(root.left, arr, x) || getPath(root.right, arr, x))
        return true;
 
    // required node does not lie either in the
    // left or right subtree of the current node
    // Thus, remove current node's value from
    // 'arr'and then return false
    arr.remove(arr.size()-1);
    return false;
}
 
// Function to get the sum of odd nodes in the
// path between any two nodes in a binary tree
static int sumOddNodes(Node root, int n1, int n2)
{
    // vector to store the path of
    // first node n1 from root
    Vector path1= new Vector();
 
    // vector to store the path of
    // second node n2 from root
    Vector path2= new Vector();
 
    getPath(root, path1, n1);
    getPath(root, path2, n2);
 
    int intersection = -1;
 
    // Get intersection point
    int i = 0, j = 0;
    while (i != path1.size() || j != path2.size()) {
 
        // Keep moving forward until no intersection
        // is found
        if (i == j && path1.get(i) == path2.get(j)) {
            i++;
            j++;
        }
        else {
            intersection = j - 1;
            break;
        }
    }
     
    int sum = 0;
     
    // calculate sum of ODD nodes from the path
    for (i = path1.size() - 1; i > intersection; i--)
        if(path1.get(i)%2!=0)
            sum += path1.get(i);
 
    for (i = intersection; i < path2.size(); i++)
        if(path2.get(i)%2!=0)
            sum += path2.get(i);
             
    return sum;        
}
 
// Driver Code
public static void main(String args[])
{
     Node root = newNode(1);
     
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.right = newNode(6);
     
    int node1 = 5;
    int node2 = 6;
     
    System.out.print(sumOddNodes(root, node1, node2));
     
}
}

Python3

# Python3 program to find sum of all odd nodes
# in the path connecting two given nodes
 
# Binary Tree node
class Node:
    def __init__(self):
        self.data = 0
        self.left = None
        self.right = None
 
# Utility function to create a
# Binary Tree node
def newNode( data):
 
    node = Node()
    node.data = data
    node.left = None
    node.right = None
     
    return node
 
# Function to check if there is a path from root
# to the given node. It also populates
# 'arr' with the given path
def getPath(root, arr, x):
 
    # if root is None
    # there is no path
    if (root == None) :
        return False
 
    # push the node's value in 'arr'
    arr.append(root.data)
 
    # if it is the required node
    # return True
    if (root.data == x) :
        return True
 
    # else check whether the required node lies
    # in the left subtree or right subtree of
    # the current node
    if (getPath(root.left, arr, x) or getPath(root.right, arr, x)) :
        return True
 
    # required node does not lie either in the
    # left or right subtree of the current node
    # Thus, remove current node's value from
    # 'arr'and then return False
    arr.pop()
    return False
 
# Function to get the sum of odd nodes in the
# path between any two nodes in a binary tree
def sumOddNodes(root, n1, n2) :
 
    # vector to store the path of
    # first node n1 from root
    path1 = []
 
    # vector to store the path of
    # second node n2 from root
    path2 = []
 
    getPath(root, path1, n1)
    getPath(root, path2, n2)
 
    intersection = -1
 
    # Get intersection point
    i = 0
    j = 0
    while (i != len(path1) or j != len(path2)):
 
        # Keep moving forward until no intersection
        # is found
        if (i == j and path1[i] == path2[j]):
            i = i + 1
            j = j + 1
         
        else :
            intersection = j - 1
            break
         
    sum = 0
     
    i = len(path1) - 1
     
    # calculate sum of ODD nodes from the path
    while ( i > intersection ):
        if(path1[i] % 2 != 0):
            sum += path1[i]
        i = i - 1
         
    i = intersection
    while ( i < len(path2) ):
        if(path2[i] % 2 != 0) :
            sum += path2[i]
        i = i + 1
             
    return sum       
 
# Driver Code
 
root = newNode(1)
     
root.left = newNode(2)
root.right = newNode(3)
root.left.left = newNode(4)
root.left.right = newNode(5)
root.right.right = newNode(6)
     
node1 = 5
node2 = 6
     
print(sumOddNodes(root, node1, node2))
 
# This code is contributed by Arnab Kundu

C#

// C# program to find sum of all odd nodes
// in the path connecting two given nodes
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Binary Tree node
public class Node
{
    public int data;
    public Node left;
    public Node right;
}
 
// Utility function to create a
// new Binary Tree node
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = null;
    node.right = null;
     
    return node;
}
 
// Function to check if there is a path from
// root to the given node. It also populates
// 'arr' with the given path
static Boolean getPath(Node root,
                       List arr, int x)
{
    // if root is null
    // there is no path
    if (root == null)
        return false;
 
    // push the node's value in 'arr'
    arr.Add(root.data);
 
    // if it is the required node
    // return true
    if (root.data == x)
        return true;
 
    // else check whether the required node lies
    // in the left subtree or right subtree of
    // the current node
    if (getPath(root.left, arr, x) ||
        getPath(root.right, arr, x))
        return true;
 
    // required node does not lie either in the
    // left or right subtree of the current node
    // Thus, Remove current node's value from
    // 'arr'and then return false
    arr.RemoveAt(arr.Count - 1);
    return false;
}
 
// Function to get the sum of odd nodes in the
// path between any two nodes in a binary tree
static int sumOddNodes(Node root, int n1, int n2)
{
    // List to store the path of
    // first node n1 from root
    List path1 = new List();
 
    // List to store the path of
    // second node n2 from root
    List path2 = new List();
 
    getPath(root, path1, n1);
    getPath(root, path2, n2);
 
    int intersection = -1;
 
    // Get intersection point
    int i = 0, j = 0;
    while (i < path1.Count || j < path2.Count)
    {
         
        // Keep moving forward until
        // no intersection is found
        if ( i == j && path1[i] == path2[j])
        {
            i++;
            j++;
        }
        else
        {
            intersection = j - 1;
            break;
        }
    }
    int sum = 0;
     
    // calculate sum of ODD nodes from the path
    for (i = path1.Count - 1; i > intersection; i--)
        if(path1[i] % 2 != 0)
            sum += path1[i];
 
    for (i = intersection; i < path2.Count; i++)
        if(path2[i] % 2 != 0)
            sum += path2[i];
             
    return sum;        
}
 
// Driver Code
public static void Main(String []args)
{
    Node root = newNode(1);
     
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.right = newNode(6);
     
    int node1 = 5;
    int node2 = 6;
     
    Console.Write(sumOddNodes(root, node1, node2));
}
}
 
// This code is contributed by Arnub Kundu

Javascript

 
    // JavaScript program to find sum of all odd nodes
    // in the path connecting two given nodes
     
    // Binary Tree node
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
 
    // Utility function to create a
    // new Binary Tree node
    function newNode(data)
    {
          let node = new Node(data);
        return node;
    }
 
    // Function to check if there is a path from root
    // to the given node. It also populates
    // 'arr' with the given path
    function getPath(root, arr, x)
    {
        // if root is null
        // there is no path
        if (root==null)
            return false;
 
        // push the node's value in 'arr'
        arr.push(root.data);
 
        // if it is the required node
        // return true
        if (root.data == x)
            return true;
 
        // else check whether the required node lies
        // in the left subtree or right subtree of
        // the current node
        if (getPath(root.left, arr, x) || getPath(root.right, arr, x))
            return true;
 
        // required node does not lie either in the
        // left or right subtree of the current node
        // Thus, remove current node's value from
        // 'arr'and then return false
        arr.pop();
        return false;
    }
 
    // Function to get the sum of odd nodes in the
    // path between any two nodes in a binary tree
    function sumOddNodes(root, n1, n2)
    {
        // vector to store the path of
        // first node n1 from root
        let path1= [];
 
        // vector to store the path of
        // second node n2 from root
        let path2= [];
 
        getPath(root, path1, n1);
        getPath(root, path2, n2);
 
        let intersection = -1;
 
        // Get intersection point
        let i = 0, j = 0;
        while (i != path1.length || j != path2.length) {
 
            // Keep moving forward until no intersection
            // is found
            if (i == j && path1[i] == path2[j]) {
                i++;
                j++;
            }
            else {
                intersection = j - 1;
                break;
            }
        }
 
        let sum = 0;
 
        // calculate sum of ODD nodes from the path
        for (i = path1.length - 1; i > intersection; i--)
            if(path1[i]%2!=0)
                sum += path1[i];
 
        for (i = intersection; i < path2.length; i++)
            if(path2[i]%2!=0)
                sum += path2[i];
 
        return sum;        
    }
     
    let root = newNode(1);
       
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.right = newNode(6);
       
    let node1 = 5;
    let node2 = 6;
       
    document.write(sumOddNodes(root, node1, node2));
 

输出：

9

时间复杂度：O(n)
辅助空间：O(n)