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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?

递归代码如下：

class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        if(root==NULL) return res;
        return pre(root,res);
    }
    vector<int> pre(TreeNode* root,vector<int>& res)
    {
        if(root)
        {
            res.push_back(root->val);
            pre(root->left,res);
            pre(root->right,res);
        }
       return res;
    }
};

迭代代码如下：

前序遍历，相当于深度搜索，可用堆栈实现，先进后出，先右后左。参考http://blog.csdn.net/sinat_24520925/article/details/45081749

代码如下：

vector<int> preorderTraversal(TreeNode* root) {
       vector<int> res;
       if(root==NULL) return res;
       stack<TreeNode*> st;
       st.push(root);
       while(!st.empty())
       {
           TreeNode *p=st.top();
           st.pop();
           res.push_back(p->val);
           if(p->right) st.push(p->right);
           if(p->left) st.push(p->left);
       }
       return res;
    }

leetcode-Binary Tree Preorder Traversal

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原文地址：http://blog.csdn.net/sinat_24520925/article/details/45603147