Draw a binary tree with seven vertices and only one leaf.

Here is a possible binary tree with seven vertices and only one leaf:

       o
      / \
     o   o
    /   / \
   o   o   o
  /
 o

Note that "o" represents a vertex and the leaf is the bottom-left vertex.

相关问题

Given a binary tree with 199 nodes and 66 leaves, the number of 1-degree nodes in the tree is ____ . A. 66 B. 67 C. 68 D. 69

Let's assume that the number of 1-degree nodes in the tree is x.

Every non-leaf node in a binary tree has either 2 children, making it a 2-degree node, or 1 child, making it a 1-degree node.

We know that the tree has 66 leaves, which are all 0-degree nodes. So, the total number of non-leaf nodes in the tree is 199 - 66 = 133.

Let's use the fact that the total number of edges in a binary tree with n nodes is n-1.

Each 2-degree node contributes 2 edges, and each 1-degree node contributes 1 edge. So, we can write an equation based on the number of edges in the tree:

2(number of 2-degree nodes) + 1(x) = 199-1

Simplifying, we get:

2(number of 2-degree nodes) + x = 198

Now, let's use the fact that the total number of leaves in a binary tree with n nodes is (n+1)/2.

We know that the tree has 66 leaves, so:

(199+1)/2 = 100

This means that there are 100 0-degree nodes in the tree.

Using the fact that the sum of the degrees of all nodes in a tree is twice the number of edges, we can write another equation:

2(number of 2-degree nodes) + 1(x) + 0(100) = 2(199-1)

Simplifying, we get:

2(number of 2-degree nodes) + x = 396

Now we have two equations:

2(number of 2-degree nodes) + x = 198

2(number of 2-degree nodes) + x = 396

Subtracting the first equation from the second, we get:

0 + 0x = 198

This is impossible, so there must be an error in the problem statement.

Therefore, the answer is unknown.

用C语言Write a function to print the leaf nodes of the binary tree.

好的，我明白了。以下是一个用C语言编写的函数，可以打印出二叉树的叶子节点：

void printLeafNodes(struct node* root) 
{ 
    if(root == NULL) 
        return; 
    if(root->left == NULL && root->right == NULL) 
    { 
        printf("%d ", root->data); 
        return; 
    } 
    if(root->left) 
        printLeafNodes(root->left); 
    if(root->right) 
        printLeafNodes(root->right); 
} 

这个函数使用递归的方法遍历二叉树，并打印出叶子节点的值。如果节点是叶子节点，则打印出节点的值，然后返回上一级。如果节点不是叶子节点，则递归处理其左右子节点。