打印二叉树顶视图中的节点 |设置 3
二叉树的顶视图是从顶部查看树时可见的节点集。给定一棵二叉树，打印它的顶视图。输出节点可以按任何顺序打印。预期时间复杂度为 O(n)
如果 x 是其水平距离处的最顶部节点，则节点 x 在输出中。节点x的左孩子水平距离等于x水平距离减1，右孩子水平距离x水平距离加1。
例子 ：
1
/ \
2 3
/ \ / \
4 5 6 7
Top view of the above binary tree is
4 2 1 3 7
1
/ \
2 3
\
4
\
5
\
6
Top view of the above binary tree is
2 1 3 6方法：
- 这里的想法是观察，如果我们尝试从树的顶部查看树，那么只会看到垂直顺序位于顶部的节点。
- 从根目录启动BFS 。维护由节点（节点 *）类型和节点到根的水平距离组成的对队列。此外，维护一个应该将节点存储在特定水平距离的地图。
- 在处理节点时，只需检查地图中该水平距离处是否有任何节点。
- 如果有任何节点，则表示无法从顶部看到该节点，请不要考虑。否则，如果在该水平距离处没有节点，请将其存储在地图中并考虑俯视图。
以下是基于上述方法的实现：
C++
// C++ program to print top
// view of binary tree
#include
using namespace std;
// Structure of binary tree
struct Node {
Node* left;
Node* right;
int data;
};
// function to create a new node
Node* newNode(int key)
{
Node* node = new Node();
node->left = node->right = NULL;
node->data = key;
return node;
}
// function should print the topView of
// the binary tree
void topView(struct Node* root)
{
// Base case
if (root == NULL) {
return;
}
// Take a temporary node
Node* temp = NULL;
// Queue to do BFS
queue > q;
// map to store node at each horizontal distance
map mp;
q.push({ root, 0 });
// BFS
while (!q.empty()) {
temp = q.front().first;
int d = q.front().second;
q.pop();
// If any node is not at that horizontal distance
// just insert that node in map and print it
if (mp.find(d) == mp.end()) {
cout << temp->data << " ";
mp[d] = temp->data;
}
// Continue for left node
if (temp->left) {
q.push({ temp->left, d - 1 });
}
// Continue for right node
if (temp->right) {
q.push({ temp->right, d + 1 });
}
}
}
// Driver Program to test above functions
int main()
{
/* Create following Binary Tree
1
/ \
2 3
\
4
\
5
\
6*/
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->right = newNode(4);
root->left->right->right = newNode(5);
root->left->right->right->right = newNode(6);
cout << "Following are nodes in top view of Binary Tree\n";
topView(root);
return 0;
} Java
// Java program to print top
// view of binary tree
import java.util.*;
class solution
{
// structure of binary tree
static class Node {
Node left;
Node right;
int data;
};
// structure of pair
static class Pair {
Node first;
int second;
Pair(Node n,int a)
{
first=n;
second=a;
}
};
// function to create a new node
static Node newNode(int key)
{
Node node = new Node();
node.left = node.right = null;
node.data = key;
return node;
}
// function should print the topView of
// the binary tree
static void topView( Node root)
{
// Base case
if (root == null) {
return;
}
// Take a temporary node
Node temp = null;
// Queue to do BFS
Queue q = new LinkedList();
// map to store node at each vertical distance
Map mp = new TreeMap();
q.add(new Pair( root, 0 ));
// BFS
while (q.size()>0) {
temp = q.peek().first;
int d = q.peek().second;
q.remove();
// If any node is not at that vertical distance
// just insert that node in map and print it
if (mp.get(d) == null) {mp.put(d, temp.data);
}
// Continue for left node
if (temp.left!=null) {
q.add(new Pair( temp.left, d - 1 ));
}
// Continue for right node
if (temp.right!=null) {
q.add(new Pair( temp.right, d + 1 ));
}
}
for(Integer data:mp.values()){
System.out.print( data + " ");
}
}
// Driver Program to test above functions
public static void main(String args[])
{
/* Create following Binary Tree
1
/ \
2 3
\
4
\
5
\
6*/
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.right = newNode(4);
root.left.right.right = newNode(5);
root.left.right.right.right = newNode(6);
System.out.println( "Following are nodes in top view of Binary Tree\n");
topView(root);
}
}
//contributed by Arnab Kundu Python3
# Python3 program to print top
# view of binary tree
# Structure of binary tree
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Function to create a new node
def newNode(key):
node = Node(key)
return node
# Function should print the topView of
# the binary tree
def topView(root):
# Base case
if (root == None):
return
# Take a temporary node
temp = None
# Queue to do BFS
q = []
# map to store node at each
# vertical distance
mp = dict()
q.append([root, 0])
# BFS
while (len(q) != 0):
temp = q[0][0]
d = q[0][1]
q.pop(0)
# If any node is not at that vertical
# distance just insert that node in
# map and print it
if d not in sorted(mp):
mp[d] = temp.data
# Continue for left node
if (temp.left):
q.append([temp.left, d - 1])
# Continue for right node
if (temp.right):
q.append([temp.right, d + 1])
for i in sorted(mp):
print(mp[i], end = ' ')
# Driver code
if __name__=='__main__':
''' Create following Binary Tree
1
/ \
2 3
\
4
\
5
\
6'''
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.right = newNode(4)
root.left.right.right = newNode(5)
root.left.right.right.right = newNode(6)
print("Following are nodes in "
"top view of Binary Tree")
topView(root)
# This code is contributed by rutvik_56C#
// C# program to print top
// view of binary tree
using System;
using System.Collections.Generic;
class GFG
{
// structure of binary tree
public class Node
{
public Node left;
public Node right;
public int data;
};
// structure of pair
public class Pair
{
public Node first;
public int second;
public Pair(Node n,int a)
{
first = n;
second = a;
}
};
// function to create a new node
static Node newNode(int key)
{
Node node = new Node();
node.left = node.right = null;
node.data = key;
return node;
}
// function should print the topView of
// the binary tree
static void topView( Node root)
{
// Base case
if (root == null)
{
return;
}
// Take a temporary node
Node temp = null;
// Queue to do BFS
Queue q = new Queue();
// map to store node at each vertical distance
Dictionary mp = new Dictionary();
q.Enqueue(new Pair( root, 0 ));
// BFS
while (q.Count>0)
{
temp = q.Peek().first;
int d = q.Peek().second;
q.Dequeue();
// If any node is not at that vertical distance
// just insert that node in map and print it
if (!mp.ContainsKey(d))
{
Console.Write( temp.data + " ");
mp.Add(d, temp.data);
}
// Continue for left node
if (temp.left != null)
{
q.Enqueue(new Pair( temp.left, d - 1 ));
}
// Continue for right node
if (temp.right != null)
{
q.Enqueue(new Pair( temp.right, d + 1 ));
}
}
}
// Driver code
public static void Main(String []args)
{
/* Create following Binary Tree
1
/ \
2 3
\
4
\
5
\
6*/
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.right = newNode(4);
root.left.right.right = newNode(5);
root.left.right.right.right = newNode(6);
Console.Write( "Following are nodes in top view of Binary Tree\n");
topView(root);
}
}
/* This code contributed by PrinciRaj1992 */ Javascript
输出：
Following are nodes in top view of Binary Tree
2 1 3 6