二叉树的前中后序遍历的模板实现

模拟递归的过程，利用栈的特点来实现遍历。
我们增加一个结构体，用来标示当前的状态：printf/go 来完成前中后序的遍历。

 struct Command {     string s;   //ptint/go两种状态     TreeNode* node;     Command(string s,TreeNode* node)     :s(s),node(node)     {} };

前序遍历

 struct Command {     string s;     TreeNode* node;     Command(string s,TreeNode* node)     :s(s),node(node)     {} };class Solution {public:    vector<int> preorderTraversal(TreeNode* root) {        vector<int >res;        if(root==NULL)        {            return res;        }        stack<Command> stack;        stack.push(Command("go",root));        while(!stack.empty())        {            Command command=stack.top();            stack.pop();            if(command.s=="print")            {                res.push_back(command.node->val);            }            else            {            if(command.node->right)            {                stack.push(Command("go",command.node->right));            }            if(command.node->left)            {                stack.push(Command("go",command.node->left));            }            stack.push(Command("print",command.node));//控制入栈与访问的先后顺序            }        }        return res;    }

中序遍历

 struct Command {     string s;     TreeNode* node;     Command(string s,TreeNode* node)     :s(s),node(node)     {} };class Solution {public:    vector<int> preorderTraversal(TreeNode* root) {        vector<int >res;        if(root==NULL)        {            return res;        }        stack<Command> stack;        stack.push(Command("go",root));        while(!stack.empty())        {            Command command=stack.top();            stack.pop();            if(command.s=="print")            {                res.push_back(command.node->val);            }            else            {            if(command.node->right)            {                stack.push(Command("go",command.node->right));            }                        stack.push(Command("print",command.node));            if(command.node->left)            {                stack.push(Command("go",command.node->left));            }            }        }        return res;    }

后序遍历

 struct Command {     string s;     TreeNode* node;     Command(string s,TreeNode* node)     :s(s),node(node)     {} };class Solution {public:    vector<int> preorderTraversal(TreeNode* root) {        vector<int >res;        if(root==NULL)        {            return res;        }        stack<Command> stack;        stack.push(Command("go",root));        while(!stack.empty())        {            Command command=stack.top();            stack.pop();            if(command.s=="print")            {                res.push_back(command.node->val);            }            else            {                                                           stack.push(Command("print",command.node));            if(command.node->right)            {                stack.push(Command("go",command.node->right));            }            if(command.node->left)            {                stack.push(Command("go",command.node->left));            }            }        }        return res;    }