[leetcode]Binary Tree Preorder Traversal

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Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1    \     2    /   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

/** * 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> preorderTraversal(TreeNode *root) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        vector<int> vi;                if(NULL == root) return vi;                TreeNode *ptr;        stack<TreeNode*> st;        st.push(root);                while(!st.empty()){            ptr = st.top(); st.pop();            vi.push_back(ptr->val);                        if(ptr->right) st.push(ptr->right);            if(ptr->left) st.push(ptr->left);        }                return vi;    }};

class Solution {public:    vector<int> preorderTraversal(TreeNode *root) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        vector<int> vi;        TreeNode *ptr = root;        stack<TreeNode*> st;        while(ptr){            while(ptr){                vi.push_back(ptr->val);                if(ptr->right){                    st.push(ptr->right);                }                ptr = ptr->left;            }            if(!st.empty()){                ptr = st.top();                st.pop();            }        }        return vi;    }};

class Solution {public:    vector<int> preorderTraversal(TreeNode *root) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        vector<int> vi;        TreeNode *ptr = root;        stack<TreeNode*> st;        st.push(root);        while(!st.empty()){            if(ptr){                vi.push_back(ptr->val);                if(ptr->right){ st.push(ptr->right);}                ptr = ptr->left;            }else{                ptr = st.top();                st.pop();            }        }        return vi;    }};

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