Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

思路：

要求递归实现，所以引入一个栈。又因为避免重复进入已访问的节点，所以用pair来记录某个节点是否被访问。

代码：

    vector<int> preorderTraversal(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        stack<pair<TreeNode*, bool> > s;
        vector<int> result;
        if(root == NULL)
            return result;
        s.push(pair<TreeNode*, bool>(root, false));
        pair<TreeNode*, bool> tPair;
        TreeNode* tNode;
        bool hasVisited;
        while(!s.empty()){
            tPair = s.top();
            tNode = tPair.first;
            hasVisited = tPair.second;
            if(hasVisited){
                s.pop();
                if(tNode->right != NULL)
                    s.push(pair<TreeNode*, bool>(tNode->right, false));
            }
            else{
                s.pop();
                s.push(pair<TreeNode*, bool>(tNode, true));
                result.push_back(tNode->val);
                if(tNode->left != NULL)
                    s.push(pair<TreeNode*, bool>(tNode->left, false));
            }
        }
        return result;
    }