Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1 \ 2 / 3
return [1,3,2]
.
Note: Recursive solution is trivial, could you do it iteratively?
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> inorderTraversal(TreeNode* root) { 13 vector<int> res; 14 inorderhelp(root,res); 15 return res; 16 17 18 } 19 void inorderhelp(TreeNode* root,vector<int> &res) 20 { 21 if(root==NULL) 22 return; 23 inorderhelp(root->left,res); 24 res.push_back(root->val); 25 inorderhelp(root->right,res); 26 return; 27 } 28 }; 29 class Solution { 30 public: 31 vector<int> inorderTraversal(TreeNode *root) 32 { 33 vector<int> res; 34 stack<pair<TreeNode* ,int>> s; 35 s.push(make_pair(root,0)); 36 while(!s.empty()) 37 { 38 TreeNode *now=s.top().first; 39 if(now==NULL) 40 s.pop(); 41 else{ 42 switch(s.top().second++) 43 { 44 case 0: 45 s.push(make_pair(now->left,0)); 46 break; 47 case 1: 48 res.push_back(now->val); 49 50 break; 51 default: 52 s.pop(); 53 s.push(make_pair(now->right,0)); 54 break; 55 } 56 57 } 58 } 59 return res; 60 } 61 62 };