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?

二叉樹前序遍歷，很簡單。

首先是遞迴方法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void trans(vector<int> &res, TreeNode* root) {
        res.push_back(root->val);
        if(root->left)  trans(res, root->left);
        if(root->right) trans(res, root->right);
        return ;
    }
    
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        if(!root)
            return res;
        trans(res, root);
        return res;
    }
};
 

接下來是迭代方法：

/**
 * Definition for a binary tree node.
 * 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> res;
        if(root == NULL)   return res;
        
        stack<TreeNode*> st;
        st.push(root);
        
        while(!st.empty()) {
            TreeNode* cur = st.top();
            st.pop();
            while(cur != NULL) {
                res.push_back(cur->val);
                if(cur->right) {
                    st.push(cur->right);
                }
                cur = cur->left;
            }
        }
        return res;
    }
};