【题目】

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? 

【代码】

/**********************************   日期：2014-11-17*   作者：SJF0115*   题号: Binary Tree Inorder Traversal*   来源：https://oj.leetcode.com/problems/binary-tree-inorder-traversal/*   结果：AC*   来源：LeetCode*   总结：**********************************/#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:    vector
         
           inorderTraversal(TreeNode *root) {        vector
          
            v; if (root == NULL){ return v; } // 根节点入栈 stack
           
             stack; TreeNode* node = root; // 遍历 while(node != NULL || !stack.empty()){ //遍历左子树 if(node != NULL){ stack.push(node); node = node->left; } else{ //左子树为空，訪问右子树 node = stack.top(); stack.pop(); v.push_back(node->val); node = node->right; } } return v; }};//按先序序列创建二叉树int CreateBTree(TreeNode* &T){ char data; //按先序次序输入二叉树中结点的值（一个字符），‘#’表示空树 cin>>data; if(data == '#'){ T = NULL; } else{ T = (TreeNode*)malloc(sizeof(TreeNode)); //生成根结点 T->val = data-'0'; //构造左子树 CreateBTree(T->left); //构造右子树 CreateBTree(T->right); } return 0;}int main() { Solution solution; TreeNode* root(0); CreateBTree(root); vector
            
              v = solution.inorderTraversal(root); for(int i = 0;i < v.size();i++){ cout<
             
              <