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?

confused what "{1,#,2,3}" means?

> read more on how binary tree is serialized on OJ.

Subscribe to see which companies asked this question

//遞歸方法
 class Solution {
 public:
	 vector<int> res;
	 vector<int> inorderTraversal(TreeNode* root) {
		 tmpFunction(root);
		 return res;
	 }

	 void tmpFunction(TreeNode* root){
		 if (root == NULL) return;
		 tmpFunction(root->left);
		 res.push_back(root->val);
		 tmpFunction(root->right);
	 }
 };

 //非遞歸方法
 class Solution {
 public:
	 vector<int> inorderTraversal(TreeNode* root) {
		 stack<TreeNode*> s;
		 vector<int> res;
		 if (root == NULL) 
			 return res;
		 TreeNode* p=root;
		 while (p!=NULL || !s.empty())
		 {
			 while (p!=NULL)
			 {
				 s.push(p);
				 p = p->left; //不斷壓入左孩子節點，直到節點為NULL
			 }
			 if (!s.empty())
			 {
				 p = s.top();
				 s.pop();
				 res.push_back(p->val);
				 p = p->right;
			 }
		 }
		 return res;
	 }
 };

LeetCode 94:Binary Tree Inorder Traversal