LeetCode--Binary Tree Preorder Traversal

Binary Tree Preorder Traversal

Total Accepted: 18022 Total
Submissions: 51784My Submissions

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?

法一，递归

class Solution {
public:
    void preOrder(TreeNode* root,vector<int>& vectorStorage)
    {
        if(root!=NULL)
        {
            vectorStorage.push_back(root->val);
            preOrder(root->left,vectorStorage);
            preOrder(root->right,vectorStorage);
        }
    }

    vector<int> preorderTraversal(TreeNode *root)
    {
        vector<int> vectorTemp;
        preOrder(root,vectorTemp);
        return vectorTemp;

    }
};

法二，迭代

vector<int> preorderTraversal(TreeNode *root)
    {
        vector<int> vectorTemp;
        //preOrder(root,vectorTemp);
        //return vectorTemp;
        if(root==NULL)return vectorTemp;
        stack<TreeNode*>S;
        TreeNode* p;
        S.push(root);
        while(!S.empty())
        {
            p=S.top();
            S.pop();
            vectorTemp.push_back(p->val);
            if(p->right!=NULL)S.push(p->right);
            if(p->left!=NULL)S.push(p->left);

        }
        return vectorTemp;
    }