代码随想录算法训练营第二十三天| 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

669. 修剪二叉搜索树

题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

解题思路：如果当前结点小于所给区间，那该节点及其左子树肯定不符合条件，返回其右子树作为上一结点子树；反之亦然。

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

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

题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

解题思路：用折半查找法，取中间值为根节点

C:

typedef struct TreeNode TreeNode;
struct TreeNode* traversal(int* nums, int left, int right) {
    if (left > right) return NULL;
    int mid = left + ((right - left) / 2);
    TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode));
    root->val=nums[mid];
    root->left = traversal(nums, left, mid - 1);
    root->right = traversal(nums, mid + 1, right);
    return root;
}
struct TreeNode* sortedArrayToBST(int* nums, int numsSize) {
    TreeNode* root = traversal(nums, 0, numsSize - 1);
    return root;
}

java：

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        return sortedArrayToBST(nums, 0, nums.length);
    }
    public TreeNode sortedArrayToBST(int[] nums, int left, int right) {
        if (left >= right) {
            return null;
        }
        if (right - left == 1) {
            return new TreeNode(nums[left]);
        }
        int mid = left + (right - left) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = sortedArrayToBST(nums, left, mid);
        root.right = sortedArrayToBST(nums, mid + 1, right);
        return root;
    }
}

538.把二叉搜索树转换为累加树

题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

解题思路：逆中序遍历

java：

class Solution {
    TreeNode pre=null;
    public TreeNode convertBST(TreeNode root) {
        if(root==null) return null;
        convertBST(root.right);
        if(pre!=null) root.val+=pre.val;
        pre=root;
        convertBST(root.left);
        return root;
    }
}