Binary Tree Preorder Traversal——经典算法的迭代求解(前序，中序，后序都在这里了)

先序遍历，用递归来做，简单的不能再简单了。代码如下：

（以下仅实现了先序遍历，中序遍历类似，后序遍历和这两个思路不一样，具体详见Binary Tree Postorder Traversal）

/**
 * 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 {
private:
      vector<int> res;
public:
    void getOrder(TreeNode* node)
    {
        if(node==NULL)
            return ;
        else
        {
            res.push_back(node->val);
            getOrder(node->left);
            getOrder(node->right);
        }
        return ;
    }
    vector<int> preorderTraversal(TreeNode* root) {
        if(root==NULL)
            return res;
        res.push_back(root->val);
        getOrder(root->left);
        getOrder(root->right);
        return res;

    }
};

但是题目要求用迭代来做。

迭代的代码如下：

/**
 * 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;
        stack<TreeNode*> stk;
        while(root!=NULL||!stk.empty())
        {
           if(root!=NULL)
           {
               res.push_back(root->val);
               stk.push(root);
               root=root->left;
           }
           else
           {
               root=stk.top()->right;
               stk.pop();
           }

        }
        return res;
    }
};

　　中序遍历迭代代码如下：

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