Binary Tree Traversal(Preorder, Inorder, Postorder )

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?
144. Binary Tree Preorder Traversal
94. Binary Tree Inorder Traversal
145. Binary Tree Postorder Traversal

遞歸：

//Recursive  preorder  Runtime: 3 ms
/**
 * Definition for a binary tree node.
 * 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) {
        vector<int> res;
        if(root==NULL){
            return res;
        }
        if(root){
            res.push_back(root->val);
        }
        if(root->left){
            vector<int> leftv=preorderTraversal(root->left);
            res.insert(res.end(),leftv.begin(),leftv.end());
        }
        if(root->right){
            vector<int> rightv=preorderTraversal(root->right);
            res.insert(res.end(),rightv.begin(),rightv.end());
        }
        return res;
    }
};

//Recursive  inorder Runtime: 0 ms
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        if(root==NULL){
            return res;
        }
        if(root->left){
            vector<int> leftv=inorderTraversal(root->left);
            res.insert(res.end(),leftv.begin(),leftv.end());
        }
        res.push_back(root->val);
        if(root->right){
            vector<int> rightv=inorderTraversal(root->right);
            res.insert(res.end(),rightv.begin(),rightv.end());
        }
        return res;
    }
};

//Recursive  postorder  Runtime: 0 ms
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        if(root==NULL){
            return res;
        }
        if(root->left){
            vector<int> leftv=postorderTraversal(root->left);
            res.insert(res.end(),leftv.begin(),leftv.end());
        }
        if(root->right){
            vector<int> rightv=postorderTraversal(root->right);
            res.insert(res.end(),rightv.begin(),rightv.end());
        }
        res.push_back(root->val);
        return res;
    }
}

迭代：

//iterative    preorder    Runtime: 0 ms
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> prestack;
        while(root||!prestack.empty()){
            while(root){
                prestack.push(root);
                res.push_back(root->val);
                root=root->left;
            }
            root=prestack.top();
            prestack.pop();
            root=root->right;
        }
        return res;
    }
};

// iterative inorder Runtime: 3 ms
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> prestack;
        while(root||!prestack.empty()){
            while(root){
                prestack.push(root);

                root=root->left;
            }
            root=prestack.top();
            res.push_back(root->val);
            prestack.pop();
            root=root->right;
        }
        return res;
    }
};

//interative postorder Runtime: 0 ms
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> prestack;
        while(root||!prestack.empty()){
            while(root){
                prestack.push(root);
                res.insert(res.begin(),root->val);
                root=root->right;
            }
            root=prestack.top();
            prestack.pop();
            root=root->left;
        }
        return res;
    }
};