1--二叉树的最近公共祖先

主要思路：

        最近祖先只有两种情况：① 自底向上，当两个目的结点分别在当前结点的左右子树时，当前结点为两个目的结点的最近祖先；② 最近祖先与其中一个目的结点相同，则另一个目的结点在目的结点的子树上；

        递归寻找目的结点，当找到目的结点后往上返回目的结点，否则返回 NULL；当一个结点在左右子树上分别找到了两个目的结点，表明这个结点是最近祖先；否则返回不为空的子树的返回结点（这时两个结点对应第 ② 种情况）；

#include <iostream>
#include <vector>
#include <stack>struct TreeNode {int val;TreeNode *left;TreeNode *right;TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};class Solution {
public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(root == NULL || root->val == p->val || root->val == q->val) return root;TreeNode* left = lowestCommonAncestor(root->left, p, q);TreeNode* right = lowestCommonAncestor(root->right, p, q);if(left != NULL && right != NULL) return root;else if( left != NULL) return left;else return right;}
};int main(int argc, char *argv[]){TreeNode *Node1 = new TreeNode(3);TreeNode *Node2 = new TreeNode(5);TreeNode *Node3 = new TreeNode(1);TreeNode *Node4 = new TreeNode(6);TreeNode *Node5 = new TreeNode(2);TreeNode *Node6 = new TreeNode(0);TreeNode *Node7 = new TreeNode(8);TreeNode *Node8 = new TreeNode(7);TreeNode *Node9 = new TreeNode(4);Node1->left = Node2;Node1->right = Node3;Node2->left = Node4;Node2->right = Node5;Node3->left = Node6;Node3->right = Node7;Node5->left = Node8;Node6->right = Node9;Solution S1;TreeNode* res = S1.lowestCommonAncestor(Node1, Node2, Node9);std::cout << res->val << std::endl;return 0;
}

2--二叉搜索树的中序后继结点

主要思路：

        如果 p 结点有右子树，则返回其右子树最左边的结点（中序遍历的定义）；

        如果 p 结点没有右子树，则从 root 结点开始寻找 p 结点的父亲结点；（根据二叉搜索树的定义，可以节省寻找的时间，只需在一边进行寻找）；

#include <iostream>
#include <vector>
#include <stack>struct TreeNode {int val;TreeNode *left;TreeNode *right;TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};class Solution {
public:TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {TreeNode *res = NULL;if(p->right != NULL){ // p的中序后继是其右子树上最左的结点（即右字数上最先返回的结点）res = p->right;while(res->left != NULL) res = res->left;return res;}// p没有右子树，从root结点开始搜索p的父亲结点while(root != NULL){if(root->val > p->val){ // p在左子树上res = root;root = root->left; // 在左子树上找到最后一个比p大的结点（中序遍历是有序的，中序后继结点表明是比p结点大）}else{root = root->right; // p在右子树上}}return res;}
};int main(int argc, char *argv[]){TreeNode *Node1 = new TreeNode(2);TreeNode *Node2 = new TreeNode(1);TreeNode *Node3 = new TreeNode(3);Node1->left = Node2;Node1->right = Node3;Solution S1;TreeNode* res = S1.inorderSuccessor(Node1, Node2);std::cout << res->val << std::endl;return 0;
}

3--二叉树的序列化与反序列化

主要思路：

        利用前序遍历或层次遍历等方式进行序列化，再根据不同遍历的方式来设计反序列化的方式，利用反序列化重构二叉树；

        下面的代码使用了层次遍历进行序列化，并通过类似层次遍历的方式进行反序列化重构二叉树；

#include <iostream>
#include <string>
#include <vector>
#include <queue>struct TreeNode {int val;TreeNode *left;TreeNode *right;TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};class Codec {
public:std::string serialize(TreeNode* root) {std::string res = "";if (root == NULL) return res;std::queue<TreeNode*> q;q.push(root);res += std::to_string(root->val) + ",";while(!q.empty()){TreeNode* tmp = q.front();q.pop();if(tmp->left != NULL){q.push(tmp->left);res += std::to_string(tmp->left->val) + ",";}else res += "#,";if(tmp->right != NULL){q.push(tmp->right);res += std::to_string(tmp->right->val) + ",";}else res += "#,";}return res;}TreeNode* deserialize(std::string data) {if (data.length() == 0) return NULL;std::vector<std::string> value;for(auto i = 0; i < data.length(); i++){std::string tmp = "";while(data[i] != ','){tmp += data[i];i++;}value.push_back(tmp);}TreeNode *res = new TreeNode(std::stoi(value[0]));std::queue<TreeNode*> q;q.push(res);int idx = 1;while(!q.empty()){TreeNode* tmp = q.front();q.pop();if(value[idx] != "#"){ // 左儿子tmp->left = new TreeNode(std::stoi(value[idx]));q.push(tmp->left);} idx++;if(value[idx] != "#"){ // 右儿子tmp->right = new TreeNode(std::stoi(value[idx]));q.push(tmp->right);} idx++;}return res;}
};void printTreeNode(TreeNode *root){if (root == NULL) return;std::queue<TreeNode*> q;q.push(root);std::cout << root->val << " ";while(!q.empty()){TreeNode* tmp = q.front();q.pop();if(tmp->left != NULL){q.push(tmp->left);std::cout << tmp->left->val << " ";}else std::cout << "null" << " ";if(tmp->right != NULL){q.push(tmp->right);std::cout << tmp->right->val << " ";}else std::cout << "null" << " ";}
}int main(int argc, char *argv[]){//root = [1,2,3,null,null,4,5]TreeNode *Node1 = new TreeNode(1);TreeNode *Node2 = new TreeNode(2);TreeNode *Node3 = new TreeNode(3);TreeNode *Node4 = new TreeNode(4);TreeNode *Node5 = new TreeNode(5);Node1->left = Node2;Node1->right = Node3;Node3->left = Node4;Node3->right = Node5;Codec S1;std::cout << S1.serialize(Node1) << std::endl;TreeNode *res = S1.deserialize(S1.serialize(Node1));printTreeNode(res);return 0;
}