【leetcode】Binary Tree Inorder Traversal

Question ： 

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.

Anwser 1 ：      

/** * Definition for binary tree * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<int> result(0);                if (root == NULL)  return result;                stack<TreeNode *> S;        TreeNode* p = root;                do        {            if (p != NULL) {                S.push(p);                p = p->left;            } else {                p = S.top();                S.pop();                                result.push_back(p->val);                p = p->right;            }        }while(!S.empty() || p != NULL);                return result;    }};

Anwser 2 ： 

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:        void dfs(TreeNode *p, vector<int> &result)    {        if(p->left != NULL)            dfs(p->left, result);                    result.push_back(p->val);                if(p->right!=NULL)            dfs(p->right, result);    }        vector<int> inorderTraversal(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        TreeNode * head = root;        vector<int> result;        result.clear();                if(head!=NULL) {            dfs(head, result);        }                    return result;    }};

Anwser 3 ： 

/** * Definition for binary tree * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<int> result;        result.clear();        stack<TreeNode *> stack_in;        stack<TreeNode *> stack_out;                if(root == NULL) return result;                stack_in.push(root);        while(!stack_in.empty())        {            TreeNode *node_in = stack_in.top();            stack_in.pop();            stack_out.push(node_in);            if(node_in->right!=NULL && node_in->left!=NULL)            {                stack_in.push(node_in->right);                stack_in.push(node_in->left);            } else if(node_in->left!=NULL && node_in->right==NULL) {                    stack_in.push(node_in->left);            } else if(node_in->left==NULL && node_in->right!=NULL) {                    stack_in.push(node_in->right);                    result.push_back(node_in->val);                    stack_out.pop();            } else {                result.push_back(node_in->val);                stack_out.pop();                while(!stack_out.empty())                {                    TreeNode * tmp = stack_out.top();                    stack_out.pop();                    result.push_back(tmp->val);                    if(tmp->right!=NULL)                        break;                }            }        }        return result;    }};

參考推薦： 

Binary Tree Inorder Traversal