[LeetCode]94.Binary Tree Inorder Traversal

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

【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 <iostream>#include <malloc.h>#include <vector>#include <stack>using namespace std;struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:    vector<int> inorderTraversal(TreeNode *root) {        vector<int> v;        if (root == NULL){            return v;        }        // 根节点入栈        stack<TreeNode*> 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，访问右子树                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<int> v = solution.inorderTraversal(root);    for(int i = 0;i < v.size();i++){        cout<<v[i]<<endl;    }}