刷题记录
- [654. 最大二叉树](#654. 最大二叉树)
- [617. 合并二叉树](#617. 合并二叉树)
- [700. 二叉搜索树中的搜索](#700. 二叉搜索树中的搜索)
- [98. 验证二叉搜索树](#98. 验证二叉搜索树)
654. 最大二叉树
本题和106. 从中序与后序遍历序列构造二叉树的思路类似,递归建树,每次找出最大值赋值给当前节点,接着递归左子树和右子树。
时间复杂度: O ( n ) O(n) O(n)
空间复杂度: O ( n ) O(n) O(n)
cpp
// c++
/**
* 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* createTree(vector<int>& nums, int left, int right){
if(left>right) return NULL;
// 记录最大值下标
int max_idx = left;
int idx = left;
// 找最大
for(; idx<=right; idx++){
if(nums[idx] > nums[max_idx]) {
max_idx = idx;
}
}
// 用最大值创建节点
TreeNode* root = new TreeNode(nums[max_idx]);
// 递归左右子树
root->left = createTree(nums, left, max_idx-1);
root->right = createTree(nums, max_idx+1, right);
return root;
}
TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
TreeNode* root = createTree(nums, 0, nums.size()-1);
return root;
}
};
617. 合并二叉树
同上题思路类似,只不过是同时遍历两个树,题目新颖,第一次遇到。
时间复杂度: O ( n ) O(n) O(n)
空间复杂度: O ( n ) O(n) O(n)
cpp
// c++
/**
* 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* Order(TreeNode* root1, TreeNode* root2){
if(!root1 && !root2) return nullptr;
TreeNode* root;
if(root1 && root2){
root = new TreeNode(root1->val + root2->val);
root->left = Order(root1->left, root2->left);
root->right = Order(root1->right, root2->right);
}else if(!root1){
root = new TreeNode(root2->val);
root->left = Order(nullptr, root2->left);
root->right = Order(nullptr, root2->right);
}else{
root = new TreeNode(root1->val);
root->left = Order(root1->left, nullptr);
root->right = Order(root1->right, nullptr);
}
return root;
}
TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
// TreeNode root;
return Order(root1, root2);
}
};
700. 二叉搜索树中的搜索
借助前序遍历来查找,找到则返回该节点。
时间复杂度: O ( n ) O(n) O(n)
空间复杂度: O ( n ) O(n) O(n)
直接前序遍历
cpp
// c++
/**
* 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:
void preOrder(TreeNode* root, int val, TreeNode* &result){
if(!root) return;
if(root->val == val){
result = root;
return;
}
// 没找到时递归左右子树,找到了就停止
if(!result)
preOrder(root->left, val, result);
if(!result)
preOrder(root->right, val, result);
}
TreeNode* searchBST(TreeNode* root, int val) {
TreeNode* result = nullptr;
preOrder(root, val, result);
return result;
}
};
借助BST性质
借助BST左小右大的性质来提高查找效率。(其实就是改一个递归左右子树的判断)
cpp
// c++
/**
* 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:
void preOrder(TreeNode* root, int val, TreeNode* &result){
if(!root) return;
if(root->val == val){
result = root;
return;
}
// 没找到时递归左右子树,找到了就停止
if(root->val > val)
preOrder(root->left, val, result);
if(root->val < val)
preOrder(root->right, val, result);
}
TreeNode* searchBST(TreeNode* root, int val) {
TreeNode* result = nullptr;
preOrder(root, val, result);
return result;
}
};
98. 验证二叉搜索树
与上题类似,这里使用中序遍历。BST的中序遍历结果是一个单调递增序列,因此在遍历时出现递减(或相等)则不是BST。
时间复杂度: O ( n ) O(n) O(n)
空间复杂度: O ( n ) O(n) O(n)
中规中矩版
cpp
// c++
/**
* 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:
void inOrder(TreeNode* root, bool & result, TreeNode* &last){
if(!root) return ;
inOrder(root->left, result, last);
if(last && last->val >= root->val) {
result = false;
return;
}
last = root;
inOrder(root->right, result, last);
}
bool isValidBST(TreeNode* root) {
bool result = true;
TreeNode* last = nullptr;
inOrder(root, result, last);
return result;
}
};
简洁版:
简化了一下参数传递,改成函数值返回。
cpp
// c++
/**
* 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:
bool inOrder(TreeNode* root, TreeNode* &last){
if(!root) return true;
bool left = inOrder(root->left, last);
if(last && last->val >= root->val) {
return false;
}
last = root;
bool right = inOrder(root->right, last);
return left && right;
}
bool isValidBST(TreeNode* root) {
TreeNode* last = nullptr;
return inOrder(root, last);
}
};