给定二叉树，打印其右下角视图。二叉树的右下角视图是从右下角访问树时可见的一组节点，返回从右到左排序的节点的值。
在右下视图中，以与右下侧成45度角的角度查看树时，只有几个节点可见，其余节点则隐藏在它们后面。
例子：
Input :
1
/ \
2 3
\ \
5 4
Output : [4, 5]
Visible nodes from the bottom right direction are 4 and 5
Input :
1
/ \
2 3
/ /
4 5
\
6
Output: [3, 6, 4]
方法 ：
- 可以使用简单的递归遍历来解决该问题。
- 通过将参数传递给所有递归调用来跟踪节点的级别。考虑树的角度为45％(如示例中所述)，因此，每当我们向左移动时，树的高度都会增加一。
- 这个想法是要跟踪最大级别并以在左子树之前访问右子树的方式遍历树。
- 到目前为止，其级别已超过最大级别的节点，请打印该节点，因为这是其级别中的最后一个节点(在左子树之前遍历右子树)。
下面是上述方法的实现：
C++
// C++ program to print bottom
// right view of binary tree
#include
using namespace std;
// A binary tree node
struct Node
{
int data;
Node *left, *right;
Node(int item)
{
data = item;
left = right = NULL;
}
};
// class to access maximum level
// by reference
struct _Max_level
{
int _max_level;
};
Node *root;
_Max_level *_max = new _Max_level();
// Recursive function to print bottom
// right view of a binary tree
void bottomRightViewUtil(Node* node, int level,
_Max_level *_max_level)
{
// Base Case
if (node == NULL)
return;
// Recur for right subtree first
bottomRightViewUtil(node->right,
level, _max_level);
// If this is the last Node of its level
if (_max_level->_max_level < level)
{
cout << node->data << " ";
_max_level->_max_level = level;
}
// Recur for left subtree
// with increased level
bottomRightViewUtil(node->left,
level + 1, _max_level);
}
// A wrapper over bottomRightViewUtil()
void bottomRightView(Node *node)
{
bottomRightViewUtil(node, 1, _max);
}
void bottomRightView()
{
bottomRightView(root);
}
// Driver Code
int main()
{
root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->right->left = new Node(5);
root->right->left->right = new Node(6);
bottomRightView();
}
// This code is contributed by Arnab Kundu Java
// Java program to print bottom
// right view of binary tree
// A binary tree node
class Node {
int data;
Node left, right;
Node(int item)
{
data = item;
left = right = null;
}
}
// class to access maximum level by reference
class Max_level {
int max_level;
}
class BinaryTree {
Node root;
Max_level max = new Max_level();
// Recursive function to print bottom
// right view of a binary tree.
void bottomRightViewUtil(Node node, int level,
Max_level max_level)
{
// Base Case
if (node == null)
return;
// Recur for right subtree first
bottomRightViewUtil(node.right,
level, max_level);
// If this is the last Node of its level
if (max_level.max_level < level) {
System.out.print(node.data + " ");
max_level.max_level = level;
}
// Recur for left subtree with increased level
bottomRightViewUtil(node.left,
level + 1, max_level);
}
void bottomRightView()
{
bottomRightView(root);
}
// A wrapper over bottomRightViewUtil()
void bottomRightView(Node node)
{
bottomRightViewUtil(node, 1, max);
}
// Driver program to test the above functions
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.right.left = new Node(5);
tree.root.right.left.right = new Node(6);
tree.bottomRightView();
}
}Python3
# Python3 program to print bottom
# right view of binary tree
# A binary tree node
class Node:
# Construct to create a newNode
def __init__(self, item):
self.data = item
self.left = None
self.right = None
maxlevel = [0]
# Recursive function to print bottom
# right view of a binary tree.
def bottomRightViewUtil(node, level, maxlevel):
# Base Case
if (node == None):
return
# Recur for right subtree first
bottomRightViewUtil(node.right, level,
maxlevel)
# If this is the last Node of its level
if (maxlevel[0] < level):
print(node.data, end = " ")
maxlevel[0] = level
# Recur for left subtree with increased level
bottomRightViewUtil(node.left, level + 1,
maxlevel)
# A wrapper over bottomRightViewUtil()
def bottomRightView(node):
bottomRightViewUtil(node, 1, maxlevel)
# Driver Code
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.right.left = Node(5)
root.right.left.right = Node(6)
bottomRightView(root)
# This code is contributed by SHUBHAMSINGH10C#
// C# program to print bottom
// right view of binary tree
using System;
// A binary tree node
class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
// class to access maximum level by reference
class Max_level
{
public int max_level;
}
class GFG
{
Node root;
Max_level max = new Max_level();
// Recursive function to print bottom
// right view of a binary tree.
void bottomRightViewUtil(Node node, int level,
Max_level max_level)
{
// Base Case
if (node == null)
return;
// Recur for right subtree first
bottomRightViewUtil(node.right,
level, max_level);
// If this is the last Node of its level
if (max_level.max_level < level)
{
Console.Write(node.data + " ");
max_level.max_level = level;
}
// Recur for left subtree with increased level
bottomRightViewUtil(node.left,
level + 1, max_level);
}
void bottomRightView()
{
bottomRightView(root);
}
// A wrapper over bottomRightViewUtil()
void bottomRightView(Node node)
{
bottomRightViewUtil(node, 1, max);
}
// Driver Code
public static void Main(String []args)
{
GFG tree = new GFG();
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.right.left = new Node(5);
tree.root.right.left.right = new Node(6);
tree.bottomRightView();
}
}
// This code is contributed by Princi Singh输出：
3 6 4
时间复杂度： O(N)，其中N是二叉树的节点数。