检查二叉树是否为满二叉树
完全二叉树定义为所有节点都具有零个或两个子节点的二叉树。相反，在一棵完全二叉树中没有节点，它有一个子节点。有关完整二叉树的更多信息，请参见此处。
例如 ：
要检查二叉树是否是完整的二叉树，我们需要测试以下情况：-
1）如果二叉树节点为NULL，那么它是一棵完整的二叉树。
2）如果二叉树节点确实有空的左右子树，那么根据定义，它是一棵满二叉树。
3）如果一个二叉树节点有左右子树，那么根据定义，它是完整二叉树的一部分。在这种情况下，递归检查左右子树本身是否也是二叉树。
4）在左右子树的所有其他组合中，二叉树不是完全二叉树。
以下是检查二叉树是否为完整二叉树的实现。
C++
// C++ program to check whether a given Binary Tree is full or not
#include
using namespace std;
/* Tree node structure */
struct Node
{
int key;
struct Node *left, *right;
};
/* Helper function that allocates a new node with the
given key and NULL left and right pointer. */
struct Node *newNode(char k)
{
struct Node *node = new Node;
node->key = k;
node->right = node->left = NULL;
return node;
}
/* This function tests if a binary tree is a full binary tree. */
bool isFullTree (struct Node* root)
{
// If empty tree
if (root == NULL)
return true;
// If leaf node
if (root->left == NULL && root->right == NULL)
return true;
// If both left and right are not NULL, and left & right subtrees
// are full
if ((root->left) && (root->right))
return (isFullTree(root->left) && isFullTree(root->right));
// We reach here when none of the above if conditions work
return false;
}
// Driver Program
int main()
{
struct Node* root = NULL;
root = newNode(10);
root->left = newNode(20);
root->right = newNode(30);
root->left->right = newNode(40);
root->left->left = newNode(50);
root->right->left = newNode(60);
root->right->right = newNode(70);
root->left->left->left = newNode(80);
root->left->left->right = newNode(90);
root->left->right->left = newNode(80);
root->left->right->right = newNode(90);
root->right->left->left = newNode(80);
root->right->left->right = newNode(90);
root->right->right->left = newNode(80);
root->right->right->right = newNode(90);
if (isFullTree(root))
cout << "The Binary Tree is full\n";
else
cout << "The Binary Tree is not full\n";
return(0);
}
// This code is contributed by shubhamsingh10 C
// C program to check whether a given Binary Tree is full or not
#include
#include
#include
/* Tree node structure */
struct Node
{
int key;
struct Node *left, *right;
};
/* Helper function that allocates a new node with the
given key and NULL left and right pointer. */
struct Node *newNode(char k)
{
struct Node *node = (struct Node*)malloc(sizeof(struct Node));
node->key = k;
node->right = node->left = NULL;
return node;
}
/* This function tests if a binary tree is a full binary tree. */
bool isFullTree (struct Node* root)
{
// If empty tree
if (root == NULL)
return true;
// If leaf node
if (root->left == NULL && root->right == NULL)
return true;
// If both left and right are not NULL, and left & right subtrees
// are full
if ((root->left) && (root->right))
return (isFullTree(root->left) && isFullTree(root->right));
// We reach here when none of the above if conditions work
return false;
}
// Driver Program
int main()
{
struct Node* root = NULL;
root = newNode(10);
root->left = newNode(20);
root->right = newNode(30);
root->left->right = newNode(40);
root->left->left = newNode(50);
root->right->left = newNode(60);
root->right->right = newNode(70);
root->left->left->left = newNode(80);
root->left->left->right = newNode(90);
root->left->right->left = newNode(80);
root->left->right->right = newNode(90);
root->right->left->left = newNode(80);
root->right->left->right = newNode(90);
root->right->right->left = newNode(80);
root->right->right->right = newNode(90);
if (isFullTree(root))
printf("The Binary Tree is full\n");
else
printf("The Binary Tree is not full\n");
return(0);
} Java
// Java program to check if binary tree is full or not
/* Tree node structure */
class Node
{
int data;
Node left, right;
Node(int item)
{
data = item;
left = right = null;
}
}
class BinaryTree
{
Node root;
/* this function checks if a binary tree is full or not */
boolean isFullTree(Node node)
{
// if empty tree
if(node == null)
return true;
// if leaf node
if(node.left == null && node.right == null )
return true;
// if both left and right subtrees are not null
// the are full
if((node.left!=null) && (node.right!=null))
return (isFullTree(node.left) && isFullTree(node.right));
// if none work
return false;
}
// Driver program
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
tree.root = new Node(10);
tree.root.left = new Node(20);
tree.root.right = new Node(30);
tree.root.left.right = new Node(40);
tree.root.left.left = new Node(50);
tree.root.right.left = new Node(60);
tree.root.left.left.left = new Node(80);
tree.root.right.right = new Node(70);
tree.root.left.left.right = new Node(90);
tree.root.left.right.left = new Node(80);
tree.root.left.right.right = new Node(90);
tree.root.right.left.left = new Node(80);
tree.root.right.left.right = new Node(90);
tree.root.right.right.left = new Node(80);
tree.root.right.right.right = new Node(90);
if(tree.isFullTree(tree.root))
System.out.print("The binary tree is full");
else
System.out.print("The binary tree is not full");
}
}
// This code is contributed by Mayank JaiswalPython3
# Python program to check whether given Binary tree is full or not
# Tree node structure
class Node:
# Constructor of the node class for creating the node
def __init__(self , key):
self.key = key
self.left = None
self.right = None
# Checks if the binary tree is full or not
def isFullTree(root):
# If empty tree
if root is None:
return True
# If leaf node
if root.left is None and root.right is None:
return True
# If both left and right subtress are not None and
# left and right subtress are full
if root.left is not None and root.right is not None:
return (isFullTree(root.left) and isFullTree(root.right))
# We reach here when none of the above if conditions work
return False
# Driver Program
root = Node(10);
root.left = Node(20);
root.right = Node(30);
root.left.right = Node(40);
root.left.left = Node(50);
root.right.left = Node(60);
root.right.right = Node(70);
root.left.left.left = Node(80);
root.left.left.right = Node(90);
root.left.right.left = Node(80);
root.left.right.right = Node(90);
root.right.left.left = Node(80);
root.right.left.right = Node(90);
root.right.right.left = Node(80);
root.right.right.right = Node(90);
if isFullTree(root):
print ("The Binary tree is full")
else:
print ("Binary tree is not full")
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)C#
// C# program to check if binary tree
// is full or not
using System;
/* Tree node structure */
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
class GFG
{
public Node root;
/* This function checks if a binary
tree is full or not */
public virtual bool isFullTree(Node node)
{
// if empty tree
if (node == null)
{
return true;
}
// if leaf node
if (node.left == null && node.right == null)
{
return true;
}
// if both left and right subtrees
// are not null they are full
if ((node.left != null) && (node.right != null))
{
return (isFullTree(node.left) &&
isFullTree(node.right));
}
// if none work
return false;
}
// Driver Code
public static void Main(string[] args)
{
GFG tree = new GFG();
tree.root = new Node(10);
tree.root.left = new Node(20);
tree.root.right = new Node(30);
tree.root.left.right = new Node(40);
tree.root.left.left = new Node(50);
tree.root.right.left = new Node(60);
tree.root.left.left.left = new Node(80);
tree.root.right.right = new Node(70);
tree.root.left.left.right = new Node(90);
tree.root.left.right.left = new Node(80);
tree.root.left.right.right = new Node(90);
tree.root.right.left.left = new Node(80);
tree.root.right.left.right = new Node(90);
tree.root.right.right.left = new Node(80);
tree.root.right.right.right = new Node(90);
if (tree.isFullTree(tree.root))
{
Console.Write("The binary tree is full");
}
else
{
Console.Write("The binary tree is not full");
}
}
}
// This code is contributed by Shrikant13Javascript
输出：
The Binary Tree is full上述代码的时间复杂度为 O(n)，其中 n 是给定二叉树中的节点数。