Create 6.3.6不平衡二叉搜索树.md

Author: fmxs

6.3.6不平衡二叉搜索树.md

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+- 问题：给定一颗不平衡二叉搜索树，左子树中的节点数比右边更多，重新组织树结构以改善其平衡性，同时保持二叉搜索树的特性
+    - 思路：
+        - 对树做一个右旋转。
+
+```cpp
+#include <iostream>
+#include <algorithm>
+using namespace std;
+
+struct TreeNode{
+    int value;
+    TreeNode* lChild;
+    TreeNode* rChild;
+    TreeNode(int val) : value(val), lChild(nullptr), rChild(nullptr) {}
+};
+
+TreeNode* RotateRight(TreeNode* oldRoot){
+    TreeNode* newRoot = oldRoot->lChild;
+    oldRoot->lChild = newRoot->rChild;
+    newRoot->rChild = oldRoot;
+    return newRoot;
+}
+
+void PreorderTraversal(TreeNode* root){
+    if(root == nullptr){
+        return;
+    }
+    cout << root->value << " ";
+    PreorderTraversal(root->lChild);
+    PreorderTraversal(root->rChild);
+}
+
+TreeNode* GetTree4(){
+    // 创建一个简单的二叉树
+    TreeNode* root = new TreeNode(6);
+    root->lChild = new TreeNode(4);
+    root->rChild = new TreeNode(7);
+    
+    root->lChild->lChild = new TreeNode(2);
+    root->lChild->rChild = new TreeNode(5);
+    
+    root->lChild->lChild->lChild = new TreeNode(1);
+    root->lChild->lChild->rChild = new TreeNode(3);
+
+    return root;
+}
+
+int main(){
+    TreeNode* root = GetTree4();
+    PreorderTraversal(root);
+    cout << endl;
+    
+    TreeNode* newRoot = RotateRight(root);
+    PreorderTraversal(newRoot);
+    return 0;
+}
+```