在 C++ 中平衡二叉搜索树
c++server side programmingprogramming
假设我们有一个二叉搜索树，我们必须找到一个具有相同节点值的平衡二叉搜索树。当且仅当每个节点的两个子树的深度相差不超过 1 时，二叉搜索树才被认为是平衡的。如果有多个结果，则返回其中任何一个。所以如果树像 −
为了解决这个问题，我们将遵循以下步骤 −
定义 inorder() 方法，这将按顺序遍历序列存储到数组中
定义构造方法()，这将取低和高 −
如果低 > high 则返回 null
mid := low + (high - low) / 2
root := new node with value arr[mid]
root 的左侧 := constrain(low, mid – 1) 且 root 的右侧 := constrain(mid + 1, high)
返回 root
从 main 方法中，调用 inorder 方法并返回 constrain(0, size of arr - 1)
示例 (C++)
让我们看下面的实现，以便更好地理解 −
#include <bits/stdc++.h>
using namespace std;
class TreeNode{
public:
int val;
TreeNode *left, *right;
TreeNode(int data){
val = data;
left = right = NULL;
}
};
void insert(TreeNode **root, int val){
queue<TreeNode*> q;
q.push(*root);
while(q.size()){
TreeNode *temp = q.front();
q.pop();
if(!temp->left){
if(val != NULL)
temp->left = new TreeNode(val);
else
temp->left = new TreeNode(0);
return;
}else{
q.push(temp->left);
}
if(!temp->right){
if(val != NULL)
temp->right = new TreeNode(val);
else
temp->right = new TreeNode(0);
return;
}else{
q.push(temp->right);
}
}
}
TreeNode *make_tree(vector<int> v){
TreeNode *root = new TreeNode(v[0]);
for(int i = 1; i<v.size(); i++){
insert(&root, v[i]);
}
return root;
}
void tree_level_trav(TreeNode*root){
if (root == NULL) return;
cout << "[";
queue<TreeNode *> q;
TreeNode *curr;
q.push(root);
q.push(NULL);
while (q.size() > 1) {
curr = q.front();
q.pop();
if (curr == NULL){
q.push(NULL);
} else {
if(curr->left)
q.push(curr->left);
if(curr->right)
q.push(curr->right);
if(curr->val == 0 || curr == NULL){
cout << "null" << ", ";
}else{
cout << curr->val << ", ";
}
}
}
cout << "]"<<endl;
}
class Solution {
public:
vector <int> arr;
void inorder(TreeNode* node){
if(!node || node->val == 0) return;
inorder(node->left);
arr.push_back(node->val);
inorder(node->right);
}
TreeNode* construct(int low, int high){
if(low > high) return NULL;
int mid = low + (high - low) / 2;
TreeNode* root = new TreeNode(arr[mid]);
root->left = construct(low, mid - 1);
root->right = construct(mid + 1, high);
return root;
}
TreeNode* balanceBST(TreeNode* root) {
inorder(root);
return construct(0, (int)arr.size() - 1);
}
};
main(){
vector<int> v = {1,NULL,2,NULL,NULL,NULL,3,NULL,NULL,NULL,NULL,NULL,NULL,NULL,4};
TreeNode *root = make_tree(v);
Solution ob;
tree_level_trav(ob.balanceBST(root));
}
输入
[1,NULL,2,NULL,NULL,NULL,3,NULL,NULL,NULL,NULL,NULL,NULL,NULL,4]
输出
[2, 1, 3, 4, ]
