二叉树先序遍历

下面是leetcode上的一道题，先序遍历二叉树。

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?

对于二叉树的遍历，很容易联想到递归，那么第一种思路就是使用递归的思想，参考代码如下：

/**
 * 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> result;
        helper(root,result);
        return result;
 
    }
    void helper(TreeNode *node,vector<int> &result){
        if (node==NULL){
            return;
        }
        result.push_back(node->val);
        helper(node->left,result);
 
        helper(node->right,result);
    }
};

我们再司考一下，先序遍历就是根节点、左子节点、右子节点，那么也可以通过ADT栈来实现。

class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) {
        vector<int> res;
        stack<TreeNode *> s;
        if (root == NULL){
            return res;
        }
        s.push(root);
        while (!s.empty()){
            TreeNode *cur = s.top();
            s.pop();
            res.push_back(cur->val);
            if (cur->right!=NULL)
                s.push(cur->right);
            if (cur->left!=NULL)
                s.push(cur->left);
        }
        return res;
    }
};