二叉树的迭代遍历

//前序1
    vector<int> preorderTraversal(TreeNode* root) {
        if(!root) return {};
        vector<int> ans;
        vector<TreeNode*> stack;
        TreeNode* cur;
        stack.push_back(root);
        //前序为中左右，先压入右子树
        while(!stack.empty()){
            cur=stack.back();
            stack.pop_back();
            ans.push_back(cur->val);
            if(cur->right){
                stack.push_back(cur->right);
            }
            if(cur->left){
                stack.push_back(cur->left);
            }
        }
        return ans;
    }

//前序2（中序改的）
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        vector<TreeNode*> stack;
        TreeNode* cur=root;
        while(cur||!stack.empty()){
            while(cur){
                ans.push_back(cur->val);
                stack.push_back(cur);
                cur=cur->left;
            }
            cur=stack.back();
            stack.pop_back();
            cur=cur->right;
        }
        return ans;
    }

//中序
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        vector<TreeNode*> stack;
        TreeNode* cur=root;
        while(cur||!stack.empty()){
            while(cur){
                stack.push_back(cur);
                cur=cur->left;
            }
            cur=stack.back();
            stack.pop_back();
            ans.push_back(cur->val);
            cur=cur->right;
        }
        return ans;
    }

//后序（左右中），将前序遍历（左右中）的左右反转，压入栈中输出
    //后序（前序1改的）
    vector<int> preorderTraversal(TreeNode* root) {
        if(!root) return {};
        vector<int> ans;
        vector<TreeNode*> stack;
        vector<int> houxu;
        TreeNode* cur;
        stack.push_back(root);
        //后序为左右中，先压入左子树
        while(!stack.empty()){
            cur=stack.back();
            stack.pop_back();
            houxu.push_back(cur->val);
            if(cur->left){
                stack.push_back(cur->left);
            }
            if(cur->right){
                stack.push_back(cur->right);
            }
        }
        for(int i=houxu.size()-1;i>=0;i--){
            ans.push_back(houxu[i]);
        }
        return ans;
    }