leetcode- Preorder/Inorder/PostOrder without Recursive

Recursive solution is trivial

方法一（麻烦）：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */     vector<int> preorderTraversal(TreeNode* root) {        vector<int> result;        if(root==NULL) return result;        stack<TreeNode*> visit;        TreeNode* p=root;        visit.push(p);        result.push_back(p->val);        while(visit.size()>0){                        while(p->left){                p=p->left;                visit.push(p);                result.push_back(p->val);            }            TreeNode* tmp=visit.top();            visit.pop();            while(!tmp->right && visit.size()>0){                tmp=visit.top();                visit.pop();            }            if(visit.size()==0&& !tmp->right) // the root has a left son only                return result;             p=tmp->right;            visit.push(p);            result.push_back(p->val);        }        return result;    }    vector<int> inorderTraversal(TreeNode* root) {        vector<int> result;        if(root==NULL) return result;        stack<TreeNode*> visit;        TreeNode* p=root;        visit.push(p);        while(visit.size()>0){                        while(p->left){                p=p->left;                visit.push(p);            }            TreeNode* tmp=visit.top();            result.push_back(tmp->val);            visit.pop();            while(!tmp->right && visit.size()>0){                tmp=visit.top();                result.push_back(tmp->val);                visit.pop();            }            if(visit.size()==0&& !tmp->right) // the root has  a left son only                return result;             p=tmp->right;            visit.push(p);        }        return result;    }    //后序遍历，稍微麻烦一些，因为要区分从左子树和从右子树回溯的状态，所以加上了一个状态    // 下面的求公共最小祖先用到后序遍历     void transform(vector<pair<TreeNode*,bool>> pairs,vector<TreeNode*> &path){        for(int i=0;i<pairs.size();i++)            path.push_back(pairs[i].first);    }    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {        if(root==NULL) return NULL;        vector<pair<TreeNode*, bool>> stack1;        vector<TreeNode*> path_p,path_q;        TreeNode *sp=root;// searchp         while(sp!=NULL|| stack1.size()>0){            while(sp!=NULL){                stack1.push_back(make_pair(sp,true));  // judge every node : p or q ? after push a new node.                if(sp==p) transform(stack1,path_p);//path_p=stack; //  when find , save the stackte(path)                 if(sp==q) transform(stack1,path_q);                if(path_p.size()>0 && path_q.size()>0 && path_p.back()==p && path_q.back()==q)  break;                 sp=sp->left;            }            if(stack1.size()>0){                sp=stack1.back().first;                bool flag=stack1.back().second;                if(flag){ // the first time on the top of stack1                    stack1.back().second=false;                    sp=sp->right;                }else{// the second time on the top of stack1, pop it from stack1                    stack1.pop_back();                    sp=NULL;                }            }        }        int length=(path_p.size()<path_q.size())?path_p.size():path_q.size();        for(int i=0;i<length;i++)            if(path_p[i]!=path_q[i])                return path_p[i-1];         return path_p[length-1];           }

方法二（更简单）：

 vector<int> postorderTraversal(TreeNode* root) {        vector<int> result;        if(root==NULL) return result;        stack<pair<TreeNode*,bool>> pairs;        pairs.push(make_pair(root,false));        bool visited;        while(!pairs.empty()){            TreeNode *tmp=pairs.top().first;            visited=pairs.top().second;            pairs.pop();            if(tmp==NULL) continue;            if(visited){                result.push_back(tmp->val);            }else{                pairs.push(make_pair(tmp,true));//(1)                pairs.push(make_pair(tmp->right,false));//(2)                pairs.push(make_pair(tmp->left,false));//(3)            }        }        return result;    }

前序和中序遍历只需要调整(1),(2),(3)处的代码，用到的基本原理是：局部有序，且每个相邻的局部有重合，则整体有序

参考：
http://zisong.me/post/suan-fa/geng-jian-dan-de-bian-li-er-cha-shu-de-fang-fa