given the shape of a binary tree shown by the figure below. if its inorder traversal sequence is { e, a, d, b, f, h, c, g }, then the node on the same level of c must be:

题目中给出了二叉树的中序遍历序列为{ e, a, d, b, f, h, c, g }，因此我们可以画出如下的二叉树：

c
/ \
/ \
/ \
b g
/ \ /
a d f
/ / \
e h i

题目要求找出与节点c同一层的节点，即深度为2的节点。根据二叉树的定义可知，c节点的深度为2，因此与之同一层的节点必然是其兄弟节点b和g。因此答案为b和g。

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.

Print the heap by tree traversal (Levelorder/Preorder, postorder, inorder traversal) anyone would be OK.

Sure! Here's an example of how to print a heap (assuming it's a binary heap) by level-order traversal using Python:

def print_heap(heap):
    """
    Prints the elements of a binary heap by level-order traversal.
    """
    if not heap:
        return

    level = 0
    nodes_in_level = 1
    index = 0

    while index < len(heap):
        print("Level", level, end=": ")

        for i in range(nodes_in_level):
            if index >= len(heap):
                break

            print(heap[index], end=" ")
            index += 1

        print()
        level += 1
        nodes_in_level *= 2

This function works by keeping track of the current level and the number of nodes in that level. It then iterates through the heap, printing out the elements in each level until it reaches the end of the heap.

You can also print a heap by preorder, postorder, or inorder traversal by modifying the traversal algorithm accordingly. For example, here's how you could print a heap by preorder traversal:

def preorder_print(heap, index=0):
    """
    Prints the elements of a binary heap by preorder traversal.
    """
    if index >= len(heap):
        return

    print(heap[index], end=" ")
    preorder_print(heap, 2*index+1)
    preorder_print(heap, 2*index+2)

This function recursively prints the current node, then traverses the left and right subtrees in preorder fashion.