[Vssue]0669.修剪二叉搜索树.md

[youngyangyang04]

https://www.programmercarl.com/0669.%E4%BF%AE%E5%89%AA%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91.html

[Du1in9]

class Solution {
    public TreeNode trimBST(TreeNode node, int low, int high) {
        if(node == null) return null;
        if(node.val < low){
            return trimBST(node.right, low, high);      // 返回右子树
        }
        if(node.val > high){
            return trimBST(node.left, low, high);       // 返回左子树
        }
        node.left = trimBST(node.left, low, high);      // 修剪右子树
        node.right = trimBST(node.right, low, high);    // 修剪右子树
        return node;
    }
}

[mockyd]

// root.val > high 和 root.val < low 的时候不直接return，而是让root等于，对于我自己来说更好理解一些

class Solution {
    public TreeNode trimBST(TreeNode root, int low, int high) {
        if (root == null) return root;

        if (root.val > high) {
            root = trimBST(root.left, low, high);
        }
        else if (root.val < low) {
            root = trimBST(root.right, low, high);
        }
        else {
            root.left = trimBST(root.left, low, high);
            root.right = trimBST(root.right, low, high);
        }

        return root;
    }
}

[gWeiXP]

佬们，为什么这样会报错啊 class Solution { public: TreeNode trimBST(TreeNode root, int low, int high) { if(root->val>=low&&root->val<=high) { if(root->left!=nullptr) root->left = trimBST(root->left, low, high); if(root->right!=nullptr) root->right = trimBST(root->right, low, high); } else { if(root->left==nullptr&&root->right==nullptr) root = nullptr; else if(root->left!=nullptr&&root->right==nullptr) root = trimBST(root->left, low, high); else if(root->left==nullptr&&root->right!=nullptr) root = trimBST(root->right, low, high); else { TreeNode* node = root->right; node = root->right; while(node->left!=nullptr) node = node->left; node->left = trimBST(root->left, low, high); root = trimBST(root->right, low, high); } } return root; } };

[mockyd]

@gWeiXP

佬们，为什么这样会报错啊 class Solution { public: TreeNode trimBST(TreeNode root, int low, int high) { if(root->val>=low&&root->val<=high) { if(root->left!=nullptr) root->left = trimBST(root->left, low, high); if(root->right!=nullptr) root->right = trimBST(root->right, low, high); } else { if(root->left==nullptr&&root->right==nullptr) root = nullptr; else if(root->left!=nullptr&&root->right==nullptr) root = trimBST(root->left, low, high); else if(root->left==nullptr&&root->right!=nullptr) root = trimBST(root->right, low, high); else { TreeNode* node = root->right; node = root->right; while(node->left!=nullptr) node = node->left; node->left = trimBST(root->left, low, high); root = trimBST(root->right, low, high); } } return root; } };

我认为是最外层的else逻辑中，对于指针的错误使用导致的内存泄漏或者数据破坏。因为我不是写c++的，所以并不是很懂，仅供参考，其实可以去问GPT

[clrlov]

为什么这道题的C++版本没有手动释放删除节点的内存呢？