Leetcode 二叉树 501 236 235 701 450
501. Find Mode in Binary Search Tree
class Solution {
private:
vector num;
void traversal(TreeNode* root){
if(root == NULL) return;
traversal(root->left);
num.push_back(root->val);
traversal(root->right);
}
public:
vector findMode(TreeNode* root) {
traversal(root);
if(num.size() == 1) return vector{num[0]};
vector res;
res.push_back(num[0]);
int maxNum = 1;
int count = 1;
for(int i=1; i= maxNum){
if(count > maxNum){
res.clear();
}
maxNum = count;
res.push_back(num[i]);
}
}
return res;
}
}; 用unordered_map（不是搜索二叉树也可以用）
class Solution {
private:
unordered_map map;
void traversal(TreeNode* root){
if(root == NULL) return;
traversal(root->left);
map[root->val]++;
traversal(root->right);
}
bool static cmp(const pair& a, const pair b){
return a.second > b.second;
} // 从大到小
public:
vector findMode(TreeNode* root) {
traversal(root);
vector res;
vector> vec(map.begin(), map.end());
sort(vec.begin(), vec.end(), cmp);
res.push_back(vec[0].first);
for(int i=1; i 236. Lowest Common Ancestor of a Binary Tree
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(root == NULL) return NULL;
if(root == p || root == q) return root;
TreeNode* left = lowestCommonAncestor(root->left, p, q);
TreeNode* right = lowestCommonAncestor(root->right, p, q);
if(left != NULL && right != NULL) return root;
if(left == NULL && right == NULL) return NULL;
return left == NULL ? right:left;
}
};这道题要从最下面开始找，当p,q分别在左右两端的时候证明此时就为公共祖先
235. Lowest Common Ancestor of a Binary Search Tree
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(root == NULL) return NULL;
if(root == p || root == q) return root;
if(p->val < q->val){
swap(p, q);
}
if(root->val < p->val && root->val > q->val) return root;
else if(root->val > p->val) return lowestCommonAncestor(root->left, p, q);
else return lowestCommonAncestor(root->right, p, q);
}
};因为可以利用bst的特点，所以从上往下找就可以了
701. Insert into a Binary Search Tree
1.递归
class Solution {
public:
TreeNode* insertIntoBST(TreeNode* root, int val) {
if(root == NULL){
TreeNode* newNode = new TreeNode(val);
return newNode;
}
if(root->val > val) root->left = insertIntoBST(root->left, val);
if(root->val < val) root->right = insertIntoBST(root->right, val);
return root;
}
};2.迭代
class Solution {
public:
TreeNode* insertIntoBST(TreeNode* root, int val) {
if(root == NULL){
TreeNode* newNode = new TreeNode(val);
return newNode;
}
TreeNode* cur = root;
TreeNode* parent = root;
while(cur != NULL){
parent = cur;
if(cur->val > val) cur = cur->left;
else cur = cur->right;
}
TreeNode* node = new TreeNode(val);
if(parent->val > val) parent->left = node;
else parent->right = node;
return root;
}
};450. Delete Node in a BST
class Solution {
private:
TreeNode* getMin(TreeNode* root){
while(root->left != NULL) root = root->left;
return root;
}
public:
TreeNode* deleteNode(TreeNode* root, int key) {
if(root == NULL) return root;
if(root->val > key) root->left = deleteNode(root->left, key);
if(root->val < key) root->right = deleteNode(root->right, key);
if(root->val == key){
if(root->left == NULL) return root->right;
if(root->right == NULL) return root->left;
if(root->left != NULL && root->right != NULL){
TreeNode* node = getMin(root->right);
root->right = deleteNode(root->right, node->val);
node->left = root->left;
node->right = root->right;
root = node;
}
}
return root;
}
};遇到删除的节点，一共会有三种情况
1.左右都没有节点，直接删掉
2.有一个节点, 用这个节点取代原来的节点
3.左右都有节点，要用左子树的最大值，或者右子树的最小值替换
注意
1.第三种情况在替换的时候，要记得删除掉原本右子树最小的值
2.因为这里要返回root, 所以root的值不能改变