标签：osi 分享图片 clu pac seq ++ 操作 one problem
An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illustrate the rotation rules.
Now given a sequence of insertions, you are supposed to tell the root of the resulting AVL tree.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (<=20) which is the total number of keys to be inserted. Then N distinct integer keys are given in the next line. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print the root of the resulting AVL tree in one line.
Sample Input 1:
5
88 70 61 96 120Sample Output 1:
70Sample Input 2:
7
88 70 61 96 120 90 65Sample Output 2:
88题目大意：写出AVL树的插入算法，并输出最终的根节点。
//这个需要了解AVL平衡条件，左右高度相差≤1，根据插入的数不算进行左旋右旋调整平衡。从来没写过，学习一下大佬的算法。
代码来自：https://www.liuchuo.net/archives/2178
#include <iostream> #include<stdio.h> using namespace std; struct node { int val; struct node *left, *right; }; node *rotateLeft(node *root) { node *t = root->right; root->right = t->left; t->left = root; return t; } node *rotateRight(node *root) { node *t = root->left;//这个左旋右旋的基本操作很厉害。 root->left = t->right; t->right = root; return t; } node *rotateLeftRight(node *root) { root->left = rotateLeft(root->left); return rotateRight(root); } node *rotateRightLeft(node *root) { root->right = rotateRight(root->right); return rotateLeft(root); } int getHeight(node *root) {//真厉害，我就写不出这种递归。 if(root == NULL) return 0; return max(getHeight(root->left), getHeight(root->right)) + 1; } node *insert(node *root, int val) {//使用指针传入，实时更新 if(root == NULL) { root = new node(); root->val = val; root->left = root->right = NULL; } else if(val < root->val) { root->left = insert(root->left, val);//递归插入，厉害 if(getHeight(root->left) - getHeight(root->right) == 2)//如果左子树过高。 root = val < root->left->val ? rotateRight(root) : rotateLeftRight(root); } else { root->right = insert(root->right, val); if(getHeight(root->left) - getHeight(root->right) == -2)//如果右子树过高。 root = val > root->right->val ? rotateLeft(root) : rotateRightLeft(root); //根据这个来判断是如果>的话，那么就是插入了右子树中的右子树。 //如果是<的话，那么就是插入了右子树中的左子树，那么就需要右左旋。 } return root;//注意最终返回的是根节点，这样才建成了二叉树~~~ } int main() { int n, val; scanf("%d", &n); node *root = NULL; for(int i = 0; i < n; i++) { scanf("%d", &val); root = insert(root, val); } printf("%d", root->val); return 0; }
1.递归完成建立与插入的操作。
2.左旋、右旋、基本操作。
3.如何判断是插入左子树还是右子树。
PAT 1066 Root of AVL Tree[AVL树][难]
标签：osi 分享图片 clu pac seq ++ 操作 one problem
原文地址：https://www.cnblogs.com/BlueBlueSea/p/9502118.html
