LeetCode Binary Tree Traversal

1. 非递归的先序遍历

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?

#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> preorderTraversal(TreeNode *root) {
        vector<int> ivec;
        stack<TreeNode*> treestk;
        if(root == NULL)
            return ivec;
        TreeNode* curr = root; 
        while(curr != NULL || !treestk.empty()) {
        	while(curr != NULL) {
        		ivec.push_back(curr->val);
        		treestk.push(curr);
        		curr = curr->left;
        	} 
        	curr = (treestk.top())->right;
        	treestk.pop();
        }
        return ivec;
    }
};

2. 非递归的中序遍历

/**
 * 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) {
        vector<int> ivec;
    	if(root == NULL)
    		return ivec;
    	stack<TreeNode*> treestk;
    	TreeNode* curr = root; 
    	while(curr != NULL || !treestk.empty()) {
    		while(curr != NULL) {
    			treestk.push(curr);
				curr = curr->left; 
    		}
    		ivec.push_back((treestk.top())->val);
    		curr = (treestk.top())->right;
    		treestk.pop();	
    	}
    }
};

3. 非递归的后序遍历

#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> postorderTraversal(TreeNode *root) {
    	vector<int> ivec;
        stack<pair<TreeNode*, bool>> treestk;
        if(root == NULL)
            return ivec;
        treestk.push(make_pair(root, false));
        while(!treestk.empty()) {
        	auto& curr = treestk.top();
        	if(curr.second) {
        		ivec.push_back((curr.first)->val);
        		treestk.pop();	 
            }
            else {
            	curr.second = true;
            	if((curr.first)->right != NULL) {
            		treestk.push(make_pair((curr.first)->right,false));	
            	}
            	if((curr.first)->left != NULL) {
					treestk.push(make_pair((curr.first)->left,false));
            	}
            } 
   		}
	}
};