给定二叉树所有层级叶节点的最大和
给定具有正节点和负节点的二叉树，任务是在给定二叉树的所有级别中找到叶节点的最大和。
例子：
Input:
4
/ \
2 -5
/ \
-1 3
Output: 2
Sum of all leaves at 0th level is 0.
Sum of all leaves at 1st level is -5.
Sum of all leaves at 2nd level is 2.
Hence maximum sum is 2.
Input:
1
/ \
2 3
/ \ \
4 5 8
/ \
6 7
Output: 13方法：解决上述问题的思路是对树进行层序遍历。遍历时，分别处理不同层次的节点。对于正在处理的每个级别，计算该级别中叶节点的总和并跟踪最大总和。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// A binary tree node has data, pointer to left child
// and a pointer to right child
struct Node {
int data;
struct Node *left, *right;
};
// Function to return the maximum sum of leaf nodes
// at any level in tree using level order traversal
int maxLeafNodesSum(struct Node* root)
{
// Base case
if (root == NULL)
return 0;
// Initialize result
int result = 0;
// Do Level order traversal keeping track
// of the number of nodes at every level
queue q;
q.push(root);
while (!q.empty()) {
// Get the size of queue when the level order
// traversal for one level finishes
int count = q.size();
// Iterate for all the nodes in the queue currently
int sum = 0;
while (count--) {
// Dequeue an node from queue
Node* temp = q.front();
q.pop();
// Add leaf node's value to current sum
if (temp->left == NULL && temp->right == NULL)
sum = sum + temp->data;
// Enqueue left and right children of
// dequeued node
if (temp->left != NULL)
q.push(temp->left);
if (temp->right != NULL)
q.push(temp->right);
}
// Update the maximum sum of leaf nodes value
result = max(sum, result);
}
return result;
}
// Helper function that allocates a new node with the
// given data and NULL left and right pointers
struct Node* newNode(int data)
{
struct Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return (node);
}
// 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(8);
root->right->right->left = newNode(6);
root->right->right->right = newNode(7);
cout << maxLeafNodesSum(root) << endl;
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// A binary tree node has data,
// pointer to left child and
// a pointer to right child
static class Node
{
int data;
Node left, right;
};
// Function to return the maximum sum
// of leaf nodes at any level in tree
// using level order traversal
static int maxLeafNodesSum(Node root)
{
// Base case
if (root == null)
return 0;
// Initialize result
int result = 0;
// Do Level order traversal keeping track
// of the number of nodes at every level
Queue q = new LinkedList<>();
q.add(root);
while (!q.isEmpty())
{
// Get the size of queue when the level order
// traversal for one level finishes
int count = q.size();
// Iterate for all the nodes
// in the queue currently
int sum = 0;
while (count-- > 0)
{
// Dequeue an node from queue
Node temp = q.peek();
q.remove();
// Add leaf node's value to current sum
if (temp.left == null &&
temp.right == null)
sum = sum + temp.data;
// Enqueue left and right children of
// dequeued node
if (temp.left != null)
q.add(temp.left);
if (temp.right != null)
q.add(temp.right);
}
// Update the maximum sum of leaf nodes value
result = Math.max(sum, result);
}
return result;
}
// Helper function that allocates a new node with the
// given data and null left and right pointers
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return (node);
}
// 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(8);
root.right.right.left = newNode(6);
root.right.right.right = newNode(7);
System.out.println(maxLeafNodesSum(root));
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 implementation of the approach
# A binary tree node has data,
# pointer to left child and
# a pointer to right child
# Helper function that allocates
# a new node with the given data
# and None left and right pointers
class newNode:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Function to return the maximum sum
# of leaf nodes at any level in tree
# using level order traversal
def maxLeafNodesSum(root):
# Base case
if (root == None):
return 0
# Initialize result
result = 0
# Do Level order traversal keeping track
# of the number of nodes at every level
q = []
q.append(root)
while(len(q)):
# Get the size of queue when the level order
# traversal for one level finishes
count = len(q)
# Iterate for all the nodes
# in the queue currently
sum = 0
while (count):
# Dequeue an node from queue
temp = q[0]
q.pop(0)
# Add leaf node's value to current sum
if (temp.left == None and
temp.right == None):
sum = sum + temp.data
# Enqueue left and right children of
# dequeued node
if (temp.left != None):
q.append(temp.left)
if (temp.right != None):
q.append(temp.right)
count -= 1
# Update the maximum sum
# of leaf nodes value
result = max(sum, result)
return result
# 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(8)
root.right.right.left = newNode(6)
root.right.right.right = newNode(7)
print(maxLeafNodesSum(root))
# This code is contributed by SHUBHAMSINGH10C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// A binary tree node has data,
// pointer to left child and
// a pointer to right child
class Node
{
public int data;
public Node left, right;
};
// Function to return the maximum sum
// of leaf nodes at any level in tree
// using level order traversal
static int maxLeafNodesSum(Node root)
{
// Base case
if (root == null)
return 0;
// Initialize result
int result = 0;
// Do Level order traversal keeping track
// of the number of nodes at every level
Queue q = new Queue();
q.Enqueue(root);
while (q.Count != 0)
{
// Get the size of queue when the level order
// traversal for one level finishes
int count = q.Count;
// Iterate for all the nodes
// in the queue currently
int sum = 0;
while (count-- > 0)
{
// Dequeue an node from queue
Node temp = q.Peek();
q.Dequeue();
// Add leaf node's value to current sum
if (temp.left == null &&
temp.right == null)
sum = sum + temp.data;
// Enqueue left and right children of
// dequeued node
if (temp.left != null)
q.Enqueue(temp.left);
if (temp.right != null)
q.Enqueue(temp.right);
}
// Update the maximum sum of leaf nodes value
result = Math.Max(sum, result);
}
return result;
}
// Helper function that allocates a new node with the
// given data and null left and right pointers
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return (node);
}
// 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(8);
root.right.right.left = newNode(6);
root.right.right.right = newNode(7);
Console.WriteLine(maxLeafNodesSum(root));
}
}
// This code is contributed by PrinciRaj1992 Javascript
输出：
13