代码随想录23|669. 修剪二叉搜索树、108.将有序数组转换为二叉搜索树 以及 538.把二叉搜索树转换为累加树

1 修剪二叉搜索树

669. 修剪二叉搜索树

迭代：

class Solution {
    //迭代法：
    //1 将节点移动到[L,H]内
    //2 处理左子树
    //3 处理右子树
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
      if(root == nullptr) return nullptr;
      //1.移动节点
      while(root != nullptr && (root->val < low || root->val > high)){
          if(root->val < low) root = root->right;
          if(root->val > high) root = root->left;
      }
      //2.处理左子树
      TreeNode* cur = root;
      while(cur != nullptr){
          while(cur->left && cur->left->val < low){
              cur->left = cur->left->right; //将不符合条件的节点删除
          } 
            cur = cur->left;//移动指针继续进行循环
      }
      //3.处理右子树
      cur=root;
      while(cur != nullptr){
          while(cur->right && cur->right->val > high) cur->right = cur->right->left;
            cur = cur->right;
      }
      return root;
    }
};

递归：

class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        //终止条件
        if(root == nullptr) return nullptr;
        //处理逻辑
        //值小往左子树找，找到目标值，将其右子树（可能有符合条件的）当作整体继续寻找，然后将找到的返回给目标值的父节点
        if(root->val > high){ 
            TreeNode* leftNode = trimBST(root->left,low,high);
            return leftNode; 
        }

        if(root->val < low){
           TreeNode* rightNode = trimBST(root->right,low,high);
           return rightNode;
        }
        root->left = trimBST(root->left,low,high);
        root->right = trimBST(root->right,low,high);
        return root;
    }
};

2 将有序数组转换为二叉搜索树

108. 将有序数组转换为二叉搜索树

参考构造二叉树（进行切割选出根节点和左右子树）

class Solution {
//左闭右闭
public:
    TreeNode* traversal(vector<int>& nums,int left,int right){
        if(left > right) return nullptr;
        //int mid = (left+right)/2;
        int mid = left + ((right-left)>>1);
        TreeNode* node = new TreeNode(nums[mid]);
        node->left = traversal(nums,left,mid-1);
        node->right = traversal(nums,mid+1,right);
        return node;
    }
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        TreeNode* root = traversal(nums,0,nums.size()-1);
        return root;
    }
};

3 把二叉搜索树转换为累加树

538. 把二叉搜索树转换为累加树

递归：

class Solution {
    //数组[1,2,33]求累加，就后序从后往前遍历
    //二叉搜索树：中序（左中右）是升序，那么就倒序遍历（右中左）
public:
    TreeNode* pre = nullptr;
    void traversal(TreeNode* cur){
        if(cur == nullptr) return;

        traversal(cur->right);
        if( pre != nullptr){
            cur->val += pre->val;
        }
        pre = cur;
        traversal(cur->left); 
    }
public:
    TreeNode* convertBST(TreeNode* root) {
        traversal(root);
        return root;

    }
};

迭代

class Solution {
    //数组[1,2,33]求累加，就后序从后往前遍历
    //二叉搜索树：中序（左中右）是升序，那么就倒序遍历（右中左）

public:
    TreeNode* convertBST(TreeNode* root) {
        stack<TreeNode*> stk;
        if(root == nullptr) return root;
        TreeNode* pre = nullptr;
        TreeNode* cur = root;
        while(cur != nullptr || !stk.empty()){
            if(cur != nullptr){
                stk.push(cur);
                cur = cur->right;
            }else{
                cur = stk.top();stk.pop();
                if(pre!=nullptr){
                    cur->val += pre->val;
                }
                pre = cur;
                cur = cur->left;
            }
        }
    return root;
    }
};