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?

非递归实现树的前序遍历。前序遍历比较简单，注意入栈的时候应该按照右孩子节点在前的顺序，这样才能保证出栈顺序左先右后。

代码如下：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) 
    {
        vector<int> vi;
        if(root==NULL)
            return vi;
        stack<TreeNode*> st;
        st.push(root);
        while(!st.empty())
        {
            root=st.top();
            st.pop();
            vi.push_back(root->val);
            if(root->right!=NULL)
                st.push(root->right);
            if(root->left!=NULL)
                st.push(root->left);
        }
    }
};