Full Binary Tree

本文介绍了一种解决无限满二叉树中节点间最短路径长度查询的问题,通过递归算法实现,适用于计算机科学领域的数据结构研究。

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题目描述

In computer science, a binary tree is a tree data structure in which each node has at most two children. Consider an infinite full binary tree (each node has two children except the leaf nodes) defined as follows. For a node labelled v its left child will be labelled 2 * v and its right child will be labelled 2 * v + 1. The root is labelled as 1.

You are given n queries of the form i, j. For each query, you have to print the length of the shortest path between node labelled i and node labelled j.

输入

First line contains n(1 ≤ n ≤ 105), the number of queries. Each query consists of two space separated integers i and j(1 ≤ i, j ≤ 109) in one line.

输出

For each query, print the required answer in one line.

样例输入

5
1 2
2 3
4 3
1024 2048
3214567 9998877

样例输出

1
2
3
1
44
#include<stdio.h>
int dps(int ,int );

int main ( )
{
	int n;
	scanf("%d",&n);
	while(n--)
	{
		int a,b,i=1,j=1,x,y,c,t;
		scanf("%d%d",&a,&b);
		while(i<=a)
		{
			i*=2;
			j++;
		}
		x=j-1; i=1; j=1;
		while(i<=b)
		{
			i*=2;
			j++;
		}
		y=j-1;
		if(x==y) printf("%d\n",dps(a,b));
		else 
		{
			if(x>y) 
			{
				c=x-y; t=c;
				while(c--) a/=2;
			}
			else 
			{
				c=y-x; t=c;
				while(c--) b/=2;
			}
			printf("%d\n",dps(a,b)+t);
		}
	}
}
int dps(int a,int b)
{
	if(a==b) return 0;
	else return dps(a/2,b/2)+2;
}

下面是用C语言实现的代码,判断一棵二叉树是否为完全二叉树。 ```c #include <stdio.h> #include <stdlib.h> #include <stdbool.h> typedef struct TreeNode { int val; struct TreeNode *left; struct TreeNode *right; } TreeNode; typedef struct Queue { TreeNode **data; int front; int rear; int size; } Queue; Queue *createQueue(int size) { Queue *q = (Queue *)malloc(sizeof(Queue)); q->data = (TreeNode **)malloc(sizeof(TreeNode *) * size); q->front = q->rear = 0; q->size = size; return q; } bool isEmpty(Queue *q) { return q->front == q->rear; } bool isFull(Queue *q) { return (q->rear + 1) % q->size == q->front; } void enqueue(Queue *q, TreeNode *node) { if (isFull(q)) { return; } q->data[q->rear] = node; q->rear = (q->rear + 1) % q->size; } TreeNode *dequeue(Queue *q) { if (isEmpty(q)) { return NULL; } TreeNode *node = q->data[q->front]; q->front = (q->front + 1) % q->size; return node; } bool isCompleteTree(TreeNode* root) { if (root == NULL) { return true; } Queue *q = createQueue(1000); bool flag = false; enqueue(q, root); while (!isEmpty(q)) { TreeNode *node = dequeue(q); if (node->left) { if (flag) { return false; } enqueue(q, node->left); } else { flag = true; } if (node->right) { if (flag) { return false; } enqueue(q, node->right); } else { flag = true; } } return true; } int main() { TreeNode *root = (TreeNode *)malloc(sizeof(TreeNode)); root->val = 1; root->left = (TreeNode *)malloc(sizeof(TreeNode)); root->left->val = 2; root->right = (TreeNode *)malloc(sizeof(TreeNode)); root->right->val = 3; root->left->left = (TreeNode *)malloc(sizeof(TreeNode)); root->left->left->val = 4; root->left->right = (TreeNode *)malloc(sizeof(TreeNode)); root->left->right->val = 5; root->right->left = (TreeNode *)malloc(sizeof(TreeNode)); root->right->left->val = 6; printf("%s\n", isCompleteTree(root) ? "true" : "false"); return 0; } ``` 代码中使用了队列来存储二叉树中的节点,判断是否为完全二叉树的方法是,从根节点开始,每层的节点必须都存在,否则后面的节点都必须是叶子节点才满足完全二叉树的定义。
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