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  • 1066 Root of AVL Tree

    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 (≤) 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 96 120
    
     

    Sample Output 1:

    70
    
     

    Sample Input 2:

    7
    88 70 61 96 120 90 65
    
     

    Sample Output 2:

    88

    题意:

      给出一组数字,要求按照给出的顺序插入到一棵平衡二叉树中,最后求这棵AVL Tree的根结点中的数值。

    思路:

      首先应该清楚,在构建的过程中四种旋转的的情况。应该先利用递归将结点插入,然后再判断是否需要旋转。

    Code:

     1 #include <bits/stdc++.h>
     2 
     3 using namespace std;
     4 
     5 typedef struct Node* node;
     6 
     7 struct Node {
     8     int val;
     9     node left;
    10     node right;
    11     Node(int v) {
    12         val = v;
    13         left = NULL;
    14         right = NULL;
    15     }
    16 };
    17 
    18 node rightRotate(node root) {
    19     node temp = root->left;
    20     root->left = temp->right;
    21     temp->right = root;
    22     return temp;
    23 }
    24 
    25 node leftRotate(node root) {
    26     node temp = root->right;
    27     root->right = temp->left;
    28     temp->left = root;
    29     return temp;
    30 }
    31 
    32 node rightLeft(node root) {
    33     root->right = rightRotate(root->right);
    34     return leftRotate(root);
    35 }
    36 
    37 node leftRight(node root) {
    38     root->left = leftRotate(root->left);
    39     return rightRotate(root);
    40 }
    41 
    42 int findHeight(node root) {
    43     if (root == NULL) return 0;
    44     int l = findHeight(root->left);
    45     int r = findHeight(root->right);
    46     return max(l, r) + 1;
    47 }
    48 
    49 void insertNode(node& root, int val) {
    50     if (root == NULL) {
    51         root = new Node(val);
    52     } else if (root->val > val) {
    53         insertNode(root->left, val);
    54         int l = findHeight(root->left);
    55         int r = findHeight(root->right);
    56         if (abs(r - l) > 1) {
    57             if (root->left->val > val) {
    58                 root = rightRotate(root);
    59             } else {
    60                 root = leftRight(root);
    61             }
    62         }
    63     } else {
    64         insertNode(root->right, val);
    65         int l = findHeight(root->left);
    66         int r = findHeight(root->right);
    67         if (abs(r - l) > 1) {
    68             if (root->right->val < val) {
    69                 root = leftRotate(root);
    70             } else {
    71                 root = rightLeft(root);
    72             }
    73         }
    74     }
    75 }
    76 
    77 void preTraveser(node root) {
    78     if (root == NULL) return;
    79     cout << root->val << " ";
    80     preTraveser(root->left);
    81     preTraveser(root->right);
    82 }
    83 
    84 int main() {
    85     int n, t;
    86     cin >> n;
    87     node root = NULL;
    88     for (int i = 0; i < n; ++i) {
    89         cin >> t;
    90         insertNode(root, t);
    91     }
    92     cout << root->val << endl;
    93     return 0;
    94 }
    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/12828354.html
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