1 BST删除节点
代码语言：javascript复制/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* getMinNode(TreeNode* root) {
while (root->left)
root = root->left;
return root;
}
TreeNode* deleteNode(TreeNode* root, int key) {
if (!root) return nullptr;
if (key < root->val)
root->left = deleteNode(root->left, key);
if (key > root->val)
root->right = deleteNode(root->right, key);
if (key == root->val) {
if (!root->left && !root->right) return nullptr;
if (!root->right) return root->left;
if (!root->left) return root->right;
auto minNode = getMinNode(root->right);
root->val = minNode->val;
root->right = deleteNode(root->right, minNode->val);
}
return root;
}
};
2 将删除节点更换到叶结点后，记住叶结点指针定点删除、
这是一个可优化的方向，不过目前我用vector存储三个树指针然后传出去，效率并没有任何提升，可能是数据结构或测试用例的问题吧，如果有高人能提高效率，还请指出
代码语言：javascript复制class Solution {
private:
bool flag = true;
public:
vector<TreeNode*> getMin(TreeNode* parent, TreeNode* root) {
if (!root->left) flag = false;
while (root->left) {
parent = root;
root = root->left;
}
vector<TreeNode*> vec;
vec.emplace_back(root);
vec.emplace_back(parent);
vec.emplace_back(root->right);
return vec;
}
TreeNode* deleteNode(TreeNode* root, int key) {
if (!root) return nullptr;
if (key < root->val)
root->left = deleteNode(root->left, key);
if (key > root->val)
root->right = deleteNode(root->right, key);
if (key == root->val) {
if (!root->left && !root->right) return nullptr;
if (!root->left) return root->right;
if (!root->right) return root->left;
auto vec = getMin(root, root->right);
root->val = vec[0]->val;
(flag ? vec[1]->left : vec[1]->right) = vec[2];
}
return root;
}
};
