二叉树中从根到叶的按位与的最大值

给定一个二叉树，任务是找到从根节点到叶节点的任何路径的按位与的最大值。

例子：

Input: Below is the given graph:
 

Output: 7
Explanation:
path 1: 15->3->5 = (15 & 3 & 5) = 1
path 2: 15->3->1 =(15 & 3 & 1) = 1
path 3: 15->7->31=(15 & 7 & 31)= 7 (maximum)
path 4: 15->7->9 = (15 & 7 & 9) =1, out of these 7 is the maximum.

Input: Below is the given graph:

Output: 6
Explanation:
Path 1: 31->3->7  =  (31 & 3 & 7)  = 3
Path 2: 31->3->1 =  (31 & 3 & 1)  = 1
Path 3: 31->15->5 =  (31 & 15 & 5)  5
Path 4: 31->15->22 =  (31 & 15 & 22)  = 6, out of these 6 is the maximum.

方法：想法是遍历从根节点到叶节点的所有路径，并计算该路径中发生的所有节点的按位与。保留一个全局变量以更新所有路径中的最大按位与值。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Initialise to update the maximum
// AND value from all the path
int maxm = 0;
 
// Node structure
struct Node {
    int val;
 
    // Left & right child of the node
    Node *left, *right;
 
    // Intialize consutructor
    Node(int x)
    {
        val = x;
        left = NULL;
        right = NULL;
    }
};
 
// Function to find the maximum value
// of Bitwise AND from root to leaf
// in a Binary tree
void maxm_Anding(Node* root, int ans)
{
    // Check if root is not null
    if (!root)
        return;
 
    if (root->left == NULL
        and root->right == NULL) {
        ans &= root->val;
 
        // Find the maxmimum AND value and
        // store in global maxm varibale
        maxm = max(ans, maxm);
 
        return;
    }
 
    // Traverse left of binary tree
    maxm_Anding(root->left,
                ans & root->val);
 
    // Traverse right of the binary tree
    maxm_Anding(root->right,
                ans & root->val);
}
 
// Driver Code
int main()
{
    // Given Tree
    Node* root = new Node(15);
    root->left = new Node(3);
    root->right = new Node(7);
    root->left->left = new Node(5);
    root->left->right = new Node(1);
    root->right->left = new Node(31);
    root->right->right = new Node(9);
 
    // Function Call
    maxm_Anding(root, root->val);
 
    // Print the maximum AND value
    cout << maxm << endl;
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Initialise to update the maximum
// AND value from all the path
static int maxm = 0;
 
// Node structure
static class Node
{
    int val;
 
    // Left & right child of the node
    Node left, right;
 
    // Intialize consutructor
    Node(int x)
    {
        val = x;
        left = null;
        right = null;
    }
};
 
// Function to find the maximum value
// of Bitwise AND from root to leaf
// in a Binary tree
static void maxm_Anding(Node root, int ans)
{
    // Check if root is not null
    if (root == null)
        return;
 
    if (root.left == null && root.right == null)
    {
        ans &= root.val;
 
        // Find the maxmimum AND value and
        // store in global maxm varibale
        maxm = Math.max(ans, maxm);
 
        return;
    }
 
    // Traverse left of binary tree
    maxm_Anding(root.left,
                ans & root.val);
 
    // Traverse right of the binary tree
    maxm_Anding(root.right,
                ans & root.val);
}
 
// Driver Code
public static void main(String[] args)
{
    // Given Tree
    Node root = new Node(15);
    root.left = new Node(3);
    root.right = new Node(7);
    root.left.left = new Node(5);
    root.left.right = new Node(1);
    root.right.left = new Node(31);
    root.right.right = new Node(9);
 
    // Function Call
    maxm_Anding(root, root.val);
 
    // Print the maximum AND value
    System.out.print(maxm + "\n");
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach
 
# Initialise to update the maximum
# AND value from all the path
maxm = 0
 
# Node structure
class Node:
     
    def __init__(self, x):
         
        self.val = x
 
        # Left & right child of the node
        self.left = None
        self.right = None
 
# Function to find the maximum value
# of Bitwise AND from root to leaf
# in a Binary tree
def maxm_Anding(root: Node, ans: int) -> None:
     
    global maxm
     
    # Check if root is not null
    if not root:
        return
 
    if (root.left is None and
        root.right is None):
        ans &= root.val
 
        # Find the maxmimum AND value and
        # store in global maxm varibale
        maxm = max(ans, maxm)
         
        return
 
    # Traverse left of binary tree
    maxm_Anding(root.left, ans & root.val)
 
    # Traverse right of the binary tree
    maxm_Anding(root.right, ans & root.val)
 
# Driver Code
if __name__ == "__main__":
 
    # Given Tree
    root = Node(15)
    root.left = Node(3)
    root.right = Node(7)
    root.left.left = Node(5)
    root.left.right = Node(1)
    root.right.left = Node(31)
    root.right.right = Node(9)
 
    # Function Call
    maxm_Anding(root, root.val)
 
    # Print the maximum AND value
    print(maxm)
 
# This code is contributed by sanjeev2552

C#

// C# program for the above approach
using System;
class GFG{
 
// Initialise to update the maximum
// AND value from all the path
static int maxm = 0;
 
// Node structure
class Node
{
    public int val;
 
    // Left & right child of the node
    public Node left, right;
 
    // Intialize consutructor
    public Node(int x)
    {
        val = x;
        left = null;
        right = null;
    }
};
 
// Function to find the maximum value
// of Bitwise AND from root to leaf
// in a Binary tree
static void maxm_Anding(Node root, int ans)
{
    // Check if root is not null
    if (root == null)
        return;
 
    if (root.left == null && root.right == null)
    {
        ans &= root.val;
 
        // Find the maxmimum AND value and
        // store in global maxm varibale
        maxm = Math.Max(ans, maxm);
 
        return;
    }
 
    // Traverse left of binary tree
    maxm_Anding(root.left,
                ans & root.val);
 
    // Traverse right of the binary tree
    maxm_Anding(root.right,
                ans & root.val);
}
 
// Driver Code
public static void Main(String[] args)
{
    // Given Tree
    Node root = new Node(15);
    root.left = new Node(3);
    root.right = new Node(7);
    root.left.left = new Node(5);
    root.left.right = new Node(1);
    root.right.left = new Node(31);
    root.right.right = new Node(9);
 
    // Function Call
    maxm_Anding(root, root.val);
 
    // Print the maximum AND value
    Console.Write(maxm + "\n");
}
}
 
// This code is contributed by sapnasingh4991

输出：

7

时间复杂度： O(N ² )
辅助空间： O(1)