给定一个二叉搜索树,同时给定最小边界L 和最大边界 R。通过修剪二叉搜索树,使得所有节点的值在[L, R]中 (R>=L) 。你可能需要改变树的根节点,所以结果应当返回修剪好的二叉搜索树的新的根节点。

错误的答案:

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

正确的答案:

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

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