[LeetCode]Binary Tree Preorder Traversal

Binary Tree Preorder Traversal

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?

不算难，使用stack即可。

/**
 * 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> result;
        TreeNode* p;
        stack<TreeNode*> mystack;
        p = root;
        if(p) mystack.push(p);
        while(!mystack.empty())
        {
            p = mystack.top();
            mystack.pop();
            result.push_back(p->val);
            if(p->right) mystack.push(p->right);
            if(p->left) mystack.push(p->left);
        }
        return result;
    }
};

 递归的方法很统一。先序，中序，后续，只要修改遍历的顺序即可。

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