leetcode: binary-tree-preorder-traversal

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题目描述：

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?

题目解析：

将一颗二叉树以前序遍历的顺序存入一个vector中。先将root根节点存入vector中，再递归左子树和右子树。和前序遍历不同之处在于需要一个函数以传引用的方式将vector传入。

AC代码：

//递归版
/**
 * 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> v;
        if(root == NULL)
            return v;
        PreOrder(root,v);
        return v;
    }
    
    void PreOrder(TreeNode* root,vector<int>& res)
    {
        if(root == NULL)
            return;
        res.push_back(root->val);
        PreOrder(root->left,res);
        PreOrder(root->right,res);
    }
};

//非递归
/**
 * 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> v;
        if(root == NULL)
            return v;
        stack<TreeNode*> s;
        s.push(root);
        while(!s.empty())
        {
            TreeNode* top = s.top();
            s.pop();
            v.push_back(top->val);
            if(top->right)
                s.push(top->right);
            if(top->left)
                s.push(top->left);
        }
        return v;
    }
};

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