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].
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> preorderTraversal(TreeNode* root) { 13 vector<int> res; 14 help(root,res); 15 return res; 16 } 17 void help(TreeNode *root,vector<int> &res) 18 { 19 if(root!=NULL) 20 { 21 res.push_back(root->val); 22 help(root->left,res); 23 help(root->right,res); 24 25 } 26 } 27 }; 28 class Solution { 29 public: 30 vector<int> preorderTraversal(TreeNode *root) 31 { 32 vector<int> res; 33 stack<TreeNode* > s; 34 s.push(root); 35 while(!s.empty()) 36 { 37 TreeNode *now=s.top(); 38 s.pop(); 39 if(now!=NULL) 40 { 41 res.push_back(now->val); 42 s.push(now->right); 43 s.push(now->left); 44 } 45 } 46 return res; 47 } 48 49 };
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