代碼隨想錄算法訓練營|第二十三天|669. 修剪二叉搜索树、108.将有序数组转换为二叉搜索树、538.把二叉搜索树转换为累加树、第六章二叉樹 - 總結。刷题心得（c++）

終止條件
1. root == NULL
  1. return root;
2. root→val < low
  1. TreeNode* right = traversal(root→right, low, high);
  2. return right;\\回傳right 是因為右樹一定比當前的root節點小，所以當前root節點小於邊界，所以將左樹全部捨棄，return 右樹。
3. root→val > high
  1. TreeNode* left = traversal(root→left, low, high);
  2. return left; \\回傳left 是因為右樹一定比當前的root節點大，所以當前root節點大於邊界，所以將右樹全部捨棄，return 左樹。
如果沒有終止條件，往左右樹去遍歷。
1. root→left = traversal(root→left, low, high);
2. root→right = traversal(root→right, low, high);
return root;

Code

class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        if(root == NULL) return root;
        if(root->val < low) {
            TreeNode* right = trimBST(root->right, low, high);
            return right;
        }
        if(root->val > high) {
            TreeNode* left = trimBST(root->left, low, high);
            return left;
        }

        root->left = trimBST(root->left, low, high);
        root->right = trimBST(root->right, low, high);

        return root;
    }
};

108.将有序数组转换为二叉搜索树 - 實作

思路

錯誤思路

找到root節點
利用左右區間使用迴圈建立左右子樹
左右子樹接到root節點

正確思路

終止條件
1. if nums.size() == 0 return NULL;
2. if nums.size() == 1 return new TreeNode(nums[0]); 已經到葉子節點了
建立root節點
1. TreeNode* root = new TreeNode(nums[ nums.size() / 2]);
建立左右區間
1. vector<int> left (nums.begin(), nums.begin() + (nums.size() / 2));
2. vector<int> right (nums.begin() + (nums.size() / 2) + 1 , nums.end()); \\要移除中間節點所以要加一

Code

錯誤代碼

class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
       if(nums.size() == 0) return NULL;
       int rootvalue = nums.size() / 2;
       printf("rootvalue %d\\n", rootvalue);
       TreeNode* root = new TreeNode(nums[rootvalue]);
       TreeNode* rootLeft = new TreeNode();
       TreeNode* rootRight = new TreeNode();
       root->left = rootLeft;
       root->right = rootRight;
       for(int i = 0; i < rootvalue; i++) {
           TreeNode* right = new TreeNode(nums[i]);
           printf("i=%d\\t left->val:%d\\t \\n", i, right->val);
           rootLeft->right = right;
           rootLeft = rootLeft->right;
       }
       for(int i = rootvalue + 1; i < nums.size(); i++) {
           TreeNode* right = new TreeNode(nums[i]);
           printf("i=%d\\t right->val:%d\\t \\n", i, right->val);
           rootRight->right = right;
           rootRight = rootRight->right;
       }       
       root->left = root->left->right;
       root->right = root->right->right;
       return root;
    }
};

正確思路

class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
       if(nums.size() == 0) return NULL;
       if(nums.size() == 1) return new TreeNode(nums[0]);
       TreeNode* root = new TreeNode(nums[nums.size() / 2]);
       
       vector<int> numsLeft(nums.begin(), nums.begin() + (nums.size() / 2));
       vector<int> numsRight(nums.begin() + (nums.size() / 2) + 1, nums.end());
       root->left = sortedArrayToBST(numsLeft);
       root->right = sortedArrayToBST(numsRight);

       return root;
    }
};

538.把二叉搜索树转换为累加树 - 實作

思路

建立一個pre Node記錄前一個節點
假設root == NULL 回傳root;
中序遍歷 → 根據題意將左右順序調換
1. 右遞迴
2. 中
  1. 假設pre != null，當前節點數值累加
  2. 向後移動，並將pre = root → 紀錄前一個節點
3. 左遞迴
return root;

Code

class Solution {
public:
    TreeNode* pre = NULL;
    TreeNode* convertBST(TreeNode* root) {
        if(root==NULL) return root;

        if(root->right) convertBST(root->right);
        if(pre != NULL) {
            root->val = pre->val + root->val;
        }
        pre = root;
        if(root->left) convertBST(root->left);

        return root;
        
    }
};