Leetcode-144. 二叉树的前序遍历

给定一个二叉树，返回它的 前序 遍历。

 示例:

输入: [1,null,2,3]
1
　
　　2
　/
3

输出: [1,2,3]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？

递归解法：

/**
 * 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 {
    vector<int> fNum;
public:
    vector<int> preorderTraversal(TreeNode* root) {
        init();
        fDfs(root);
        return fNum;
    }
    void fDfs(TreeNode* root)
    {
        if (root!= NULL)
        {
            fNum.push_back(root->val);
            fDfs(root->left);  //遍历左子树
            fDfs(root->right);  //遍历右子树
        }
    }
    void init()
    {
        fNum.clear();
    }
};

迭代算法：

/**
 * 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 {
    vector<int> num;
    stack<TreeNode* > fNum; 
public:
    vector<int> preorderTraversal(TreeNode* root) {
        init();
        bfs(root);
        return num;
    }
    void bfs(TreeNode* root)
    {
        while (root!= NULL|| ! fNum.empty())
        {
            while (root!= NULL)
            {
                fNum.push(root);
                num.push_back(root->val);
                root = root->left;
            }
            root = fNum.top();
            fNum.pop();
            root = root->right;
        }
    }
    void init()
    {
        num.clear();
        if(! fNum.empty()) fNum.pop();
    }
};