LeetCode94——Binary Tree Inorder Traversal

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.

实现：

class Solution {public:    vector<int> inorderTraversal(TreeNode* root) {    vector<int> val;    vector<TreeNode*> record;    vector<TreeNode*> traverse;    if (root == NULL) return val;    record.push_back(root);    while (record.size() > 0)    {        TreeNode *node = record.back();        if ((node->left && std::find(traverse.begin(), traverse.end(), node->left) != traverse.end()) || node->left == NULL) {            val.push_back(node->val);            record.pop_back();            traverse.push_back(node);            if (node->right) {                record.push_back(node->right);            }        }        else if (node->left)        {            record.push_back(node->left);            continue;        }    }    return val;}};