Binary tree Traversals

recursive很容易写binary tree
iterative后来再补上：DFS vs BFS是最好的queue vs stack 实例

1, Depth first traversals:
inorder (left, root, right)
preorder (root, left, right)
postorder (left, right, root)
The root postion decides the order.
For example：
Inorder:

    void helper(TreeNode* root, vector<int>&res)
    {
        if(root == NULL) return;
        helper(root->left, res);
        res.push_back(root->val);
        helper(root->right, res);
    }

Preorder:

    void helper(TreeNode* root, vector<int>&res)
    {
        if(root == NULL) return;
        res.push_back(root->val);
        helper(root->left, res);
        helper(root->right, res);
    }

PostOrder:

    void helper(TreeNode* root, vector<int>&res)
    {
        if(root == NULL) return;
        helper(root->left, res);
        helper(root->right, res);
        res.push_back(root->val);
    }

2, Breadth first traversal:
level order:

/**
 * 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<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>>res;
        if(root == NULL) return res;
        
        queue<TreeNode*>q;
        q.push(root);
        while(!q.empty())
        {
            vector<int>row;
            int size = q.size();
            
            for(int i = 0; i < size; i++)
            {
                TreeNode* cur = q.front();
                q.pop();
                row.push_back(cur->val);
                if(cur->left) q.push(cur->left);
                if(cur->right) q.push(cur->right);
            }
            res.push_back(row);
                
        }
        
        return res;
    }
};