给定一棵二叉树，任务是在树的每一层打印除最右边的所有节点。根被认为是第 0 层，任何级别的最右边的节点都被认为是位置 0 的节点。
例子：
Input:
1
/ \
2 3
/ \ \
4 5 6
/ \
7 8
/ \
9 10
Output:
2
4 5
7
9
Input:
1
/ \
2 3
\ \
4 5
Output:
2
4方法：要逐层打印节点，使用层序遍历。该想法基于逐行打印级别顺序遍历。为此，逐层遍历节点，如果级别顺序队列中的节点是最后一个节点，则该节点将是最右边的节点，并且不打印该节点。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Structure of the tree node
struct Node {
int data;
Node *left, *right;
};
// Utility method to create a node
struct Node* newNode(int data)
{
struct Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Function to print all the nodes
// except the rightmost in every level
// of the given binary tree
// with level order traversal
void excluderightmost(Node* root)
{
// Base Case
if (root == NULL)
return;
// Create an empty queue for level
// order traversal
queue q;
// Enqueue root
q.push(root);
while (1) {
// nodeCount (queue size) indicates
// number of nodes at current level.
int nodeCount = q.size();
if (nodeCount == 0)
break;
// Dequeue all nodes of current level
// and Enqueue all nodes of next level
while (nodeCount > 0) {
Node* node = q.front();
// If node is not rightmost print
if (nodeCount != 1)
cout << node->data << " ";
q.pop();
if (node->left != NULL)
q.push(node->left);
if (node->right != NULL)
q.push(node->right);
nodeCount--;
}
cout << "\n";
}
}
// 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->left = newNode(6);
root->right->right = newNode(7);
root->left->right->left = newNode(8);
root->left->right->right = newNode(9);
root->left->right->right->right = newNode(10);
excluderightmost(root);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class Sol {
// Structure of the tree node
static class Node {
int data;
Node left, right;
};
// Utility method to create a node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return (node);
}
// Function to print all the nodes
// except the rightmost in every level
// of the given binary tree
// with level order traversal
static void excluderightmost(Node root)
{
// Base Case
if (root == null)
return;
// Create an empty queue for level
// order traversal
Queue q = new LinkedList();
// Enqueue root
q.add(root);
while (true) {
// nodeCount (queue size) indicates
// number of nodes at current level.
int nodeCount = q.size();
if (nodeCount == 0)
break;
// Dequeue all nodes of current level
// and Enqueue all nodes of next level
while (nodeCount > 0) {
Node node = q.peek();
// If node is not rightmost print
if (nodeCount != 1)
System.out.print(node.data + " ");
q.remove();
if (node.left != null)
q.add(node.left);
if (node.right != null)
q.add(node.right);
nodeCount--;
}
System.out.println();
}
}
// 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.left = newNode(6);
root.right.right = newNode(7);
root.left.right.left = newNode(8);
root.left.right.right = newNode(9);
root.left.right.right.right = newNode(10);
excluderightmost(root);
}
} Python
# Python implementation of the approach
from collections import deque
# Structure of the tree node
class Node:
def __init__(self):
self.data = 0
self.left = None
self.right = None
# Utility method to create a node
def newNode(data: int) -> Node:
node = Node()
node.data = data
node.left = None
node.right = None
return node
# Function to print all the nodes
# except the rightmost in every level
# of the given binary tree
# with level order traversal
def excluderightmost(root: Node):
# Base Case
if root is None:
return
# Create an empty queue for level
# order traversal
q = deque()
# Enqueue root
q.append(root)
while 1:
# nodeCount (queue size) indicates
# number of nodes at current level
nodeCount = len(q)
if nodeCount == 0:
break
# Dequeue all nodes of current level
# and Enqueue all nodes of next level
while nodeCount > 0:
node = q[0]
# If Node is not right most print
if nodeCount != 1:
print(node.data, end =" ")
q.popleft()
if node.left is not None:
q.append(node.left)
if node.right is not None:
q.append(node.right)
nodeCount -= 1
print()
# Driver Code
if __name__ == "__main__":
root = Node()
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.right.left = newNode(8)
root.left.right.right = newNode(9)
root.left.right.right.right = newNode(10)
excluderightmost(root)C#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG {
// Structure of the tree node
public class Node {
public int data;
public Node left, right;
};
// Utility method to create a node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return (node);
}
// Function to print all the nodes
// except the rightmost in every level
// of the given binary tree
// with level order traversal
static void excluderightmost(Node root)
{
// Base Case
if (root == null)
return;
// Create an empty queue for level
// order traversal
Queue q = new Queue();
// Enqueue root
q.Enqueue(root);
while (true) {
// nodeCount (queue size) indicates
// number of nodes at current level.
int nodeCount = q.Count;
if (nodeCount == 0)
break;
// Dequeue all nodes of current level
// and Enqueue all nodes of next level
while (nodeCount > 0) {
Node node = q.Peek();
// if Node is not right most print
if (nodeCount != 1)
Console.Write(node.data + " ");
q.Dequeue();
if (node.left != null)
q.Enqueue(node.left);
if (node.right != null)
q.Enqueue(node.right);
nodeCount--;
}
Console.WriteLine();
}
}
// 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.left = newNode(6);
root.right.right = newNode(7);
root.left.right.left = newNode(8);
root.left.right.right = newNode(9);
root.left.right.right.right = newNode(10);
excluderightmost(root);
}
} 输出：
2
4 5 6
8
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