给定二叉树的最长直线路径长度

给定一棵二叉树，任务是找到给定二叉树的最长直线路径的长度。

Straight Path is defined as the path that starts from any node and ends at another node in the tree such that the direction of traversal from the source node to the destination node always remains the same i.e., either left or right, without any change in direction that is left->left ->left or right->right->right direction. 

例子：

Input:

Output: 2
Explanation:
The path shown in green is the longest straight path from 4 to 6 which is of length 2. 
 

Input:

Output: 3
Explanation:
The path shown in green is the longest straight path from 5 to 0 which is of length 3. 
 

方法：想法是使用后序遍历。请按照以下步骤解决问题：

1. 对于每个节点，检查当前节点的方向(向左或向右)，并检查其子节点的哪个方向为其下方提供了该节点的最长长度。
2. 如果当前节点的方向和给出最长长度的子节点的方向不同，则保存该子节点的结果并将另一个子节点的长度传递给其父节点。
3. 使用上述步骤在每个节点找到最长的直线路径并保存结果以打印所有直线路径中的最大值。
4. 经过以上步骤打印最大路径。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Structure of a Tree node
struct Node {
    int key;
    struct Node *left, *right;
};
 
// Function to create a new node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return (temp);
}
 
// Function to find the longest
// straight path in a tree
int findPath(Node* root, char name,
             int& max_v)
{
    // Base Case
    if (root == NULL) {
        return 0;
    }
 
    // Recursive call on left child
    int left = findPath(root->left,
                        'l', max_v);
 
    // Recursive call on right child
    int right = findPath(root->right,
                         'r', max_v);
 
    // Return the maximum straight
    // path possible from current node
    if (name == 't') {
        return max(left, right);
    }
 
    // Leaf node
    if (left == 0 && right == 0) {
        return 1;
    }
 
    // Executes when either of the
    // child is present or both
    else {
 
        // Pass the longest value from
        // either direction
        if (left < right) {
            if (name == 'r')
                return 1 + right;
 
            else {
                max_v = max(max_v, right);
                return 1 + left;
            }
        }
        else {
            if (name == 'l')
                return 1 + left;
 
            else {
                max_v = max(max_v, left);
                return 1 + right;
            }
        }
    }
    return 0;
}
 
// Driver Code
int main()
{
 
    // Given Tree
    Node* root = newNode(3);
 
    root->left = newNode(3);
    root->right = newNode(3);
 
    root->left->right = newNode(2);
    root->right->left = newNode(4);
    root->right->left->left = newNode(4);
 
    int max_v = max(
        findPath(root, 't', max_v),
        max_v);
 
    // Print the maximum length
    cout << max_v << "\n";
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
static int max_v;
 
// Structure of a Tree node
static class Node
{
    int key;
    Node left, right;
};
 
// Function to create a new node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    temp.left = temp.right = null;
    return (temp);
}
 
// Function to find the longest
// straight path in a tree
static int findPath(Node root, char name)
{
     
    // Base Case
    if (root == null)
    {
        return 0;
    }
 
    // Recursive call on left child
    int left = findPath(root.left, 'l');
 
    // Recursive call on right child
    int right = findPath(root.right, 'r');
 
    // Return the maximum straight
    // path possible from current node
    if (name == 't')
    {
        return Math.max(left, right);
    }
 
    // Leaf node
    if (left == 0 && right == 0)
    {
        return 1;
    }
 
    // Executes when either of the
    // child is present or both
    else
    {
         
        // Pass the longest value from
        // either direction
        if (left < right)
        {
            if (name == 'r')
                return 1 + right;
            else
            {
                max_v = Math.max(max_v, right);
                return 1 + left;
            }
        }
        else
        {
            if (name == 'l')
                return 1 + left;
            else
            {
                max_v = Math.max(max_v, left);
                return 1 + right;
            }
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given Tree
    Node root = newNode(3);
 
    root.left = newNode(3);
    root.right = newNode(3);
    root.left.right = newNode(2);
    root.right.left = newNode(4);
    root.right.left.left = newNode(4);
 
    max_v = Math.max(findPath(root, 't'),
                     max_v);
 
    // Print the maximum length
    System.out.print(max_v+ "\n");
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
max_v = 0
 
# Structure of a Tree node
class newNode:
     
    def __init__(self, key):
         
        self.key = key
        self.left = None
        self.right = None
 
# Function to find the longest
# straight path in a tree
def findPath(root, name):
     
    global max_v
     
    # Base Case
    if (root == None):
        return 0
 
    # Recursive call on left child
    left = findPath(root.left, 'l')
 
    # Recursive call on right child
    right = findPath(root.right, 'r')
     
    # Return the maximum straight
    # path possible from current node
    if (name == 't'):
        return max(left, right)
 
    # Leaf node
    if (left == 0 and right == 0):
        return 1
 
    # Executes when either of the
    # child is present or both
    else:
         
        # Pass the longest value from
        # either direction
        if (left < right):
            if (name == 'r'):
                return 1 + right
            else:
                max_v = max(max_v, right)
                return 1 + left
        else:
            if (name == 'l'):
                return 1 + left
            else:
                max_v = max(max_v, left)
                return 1 + right
                 
    return 0
 
def helper(root):
     
    global max_v
    temp = max(findPath(root, 't'), max_v)
    print(temp)
 
# Driver Code
if __name__ == '__main__':
     
    # Given Tree
    root = newNode(3)
    root.left = newNode(3)
    root.right = newNode(3)
    root.left.right = newNode(2)
    root.right.left = newNode(4)
    root.right.left.left = newNode(4)
     
    helper(root)
 
# This code is contributed by ipg2016107

C#

// C# program for
// the above approach
using System;
class GFG{
     
static int max_v;
 
// Structure of a Tree node
public class Node
{
  public int key;
  public Node left, right;
};
 
// Function to create a new node
static Node newNode(int key)
{
  Node temp = new Node();
  temp.key = key;
  temp.left = temp.right = null;
  return (temp);
}
 
// Function to find the longest
// straight path in a tree
static int findPath(Node root,
                    char name)
{
  // Base Case
  if (root == null)
  {
    return 0;
  }
 
  // Recursive call on left child
  int left = findPath(root.left, 'l');
 
  // Recursive call on right child
  int right = findPath(root.right, 'r');
 
  // Return the maximum straight
  // path possible from current node
  if (name == 't')
  {
    return Math.Max(left, right);
  }
 
  // Leaf node
  if (left == 0 && right == 0)
  {
    return 1;
  }
 
  // Executes when either of the
  // child is present or both
  else
  {
    // Pass the longest value from
    // either direction
    if (left < right)
    {
      if (name == 'r')
        return 1 + right;
      else
      {
        max_v = Math.Max(max_v, right);
        return 1 + left;
      }
    }
    else
    {
      if (name == 'l')
        return 1 + left;
      else
      {
        max_v = Math.Max(max_v, left);
        return 1 + right;
      }
    }
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given Tree
  Node root = newNode(3);
 
  root.left = newNode(3);
  root.right = newNode(3);
  root.left.right = newNode(2);
  root.right.left = newNode(4);
  root.right.left.left = newNode(4);
 
  max_v = Math.Max(findPath(root, 't'), max_v);
 
  // Print the maximum length
  Console.Write(max_v + "\n");
}
}
 
// This code is contributed by 29AjayKumar

输出：

2

时间复杂度： O(N)
辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live