606. 根据二叉树创建字符串
要点：前序遍历，当左子树为空时，右结点有数字时要给左边加括号
class Solution {
public:string tree2str(TreeNode* root) {string s;//创建一个字符串if(root==nullptr){return s;}s+=to_string(root->val);//保存当前结点，转换为字符串if(root->left)//左子树不为空，继续往左子树走{s+="(";s+=tree2str(root->left);s+=")";}else if(root->right)//当左子树为空右子树不为空{s+="()";}if(root->right)//遍历右边{s+="(";s+=tree2str(root->right);s+=")";}//不需要加（）return s;}
};102. 二叉树的层序遍历
层次遍历想到用队列将一层一层的数据存入在拿出来
class Solution {
public:vector<vector<int>> levelOrder(TreeNode* root) {vector<vector<int>> vv;queue<TreeNode*> q;if(root){q.push(root);//存入第一个结点}while(!q.empty())//判空{vector<int> v;int lenght=q.size();for(size_t i = 0;i<lenght;i++)//用一个值去控制输出的层次{TreeNode* front = q.front();//将存入的数提取v.push_back(front->val);//给一维数组赋值q.pop();//弹出if(front->left)//当有左值存入q.push(front->left);if(front->right)//当有右值存入q.push(front->right);}vv.push_back(v);//将该一维数组存入，二维数组中}return vv;}
};107. 二叉树的层序遍历 II
和上一道题相反，只需要将结果反转最后的二维数组就可以了reverse(开始的迭代器和结束的迭代器)
class Solution {
public:vector<vector<int>> levelOrderBottom(TreeNode* root){vector<vector<int>> vv;queue<TreeNode*> q;if(root){q.push(root);//存入第一个结点}while(!q.empty())//判空{vector<int> v;int lenght=q.size();for(size_t i = 0;i<lenght;i++)//用一个值去控制输出的层次{TreeNode* front = q.front();//将存入的数提取v.push_back(front->val);//给一维数组赋值q.pop();//弹出if(front->left)//当有左值存入q.push(front->left);if(front->right)//当有右值存入q.push(front->right);}vv.push_back(v);//将该一维数组存入，二维数组中}reverse(vv.begin(),vv.end());return vv;}
};236. 二叉树的最近公共祖先
分析：这里是二叉搜索树只要知道左边和右边中间的就是公共祖先
class Solution {
public:bool Find(TreeNode* root, TreeNode* p){if(root==nullptr)return false;return root==p||Find(root->left,p)||Find(root->right,p);}TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(root==NULL)return NULL;if(root==p||root==q)return root;bool pInleft,pInright,qInleft,qInright;//这里代表pq在左边还是右边pInleft=Find(root->left,p);//这里也是递归，寻找p在那边pInright=!pInleft;qInleft=Find(root->left,q);//这里也是递归，寻找q在那边qInright=!qInleft;if((pInleft && qInright)||((qInleft && pInright))){return root;//当pq在左右两边时root就是祖先}if(qInleft && pInleft)//当pq都在左边{return lowestCommonAncestor(root->left,p,q);}if(qInright && pInright)//当pq都在右边{return lowestCommonAncestor(root->right,p,q);}return NULL;}
};
二叉搜索树与双向链表_牛客题霸_牛客网 (nowcoder.com)
class Solution {
public:void convertList(TreeNode* cur,TreeNode*& prev)//这里为什么要加&，递归到最后改变cur并没有改变所以要加&，才知道用同一个{if(cur==NULL){return;}convertList(cur->left,prev);cur->left=prev;if(prev)prev->right = cur;prev =cur;convertList(cur->right,prev);}TreeNode* Convert(TreeNode* pRootOfTree) {TreeNode* prev = NULL;convertList(pRootOfTree,prev);TreeNode* head = pRootOfTree;while(head && head->left){head=head->left;}return head;}
};
