leetcode 114. Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place.
For example, given the following tree:
1
/
2 5
/ \
3 4 6
The flattened tree should look like:
1

2

3

4

5

6
大意：将二叉树展开为链表
方法：有两种展开的方法（DFS的题目都可以用递归和迭代的方法）

方法一（递归）：

class Solution {
public:
    void flatten(TreeNode* root) {
        if(!root) return;
        if(root->left) flatten(root->left);
        if(root->right) flatten(root->right);
        TreeNode *temp=root->right;
        root->right=root->left;
        root->left=NULL;//断开左结点，才能连上右结点
        while(root->right) //返回上一个父节点，继续
            root=root->right;
        root->right=temp;  
    }
};

法二（迭代：本质是前序遍历（当然其他也行））：

class Solution {
public:
    void flatten(TreeNode* root) {
        if (!root) return;
        stack<TreeNode*> s;
        s.push(root);
        while (!s.empty()) {
            TreeNode *t = s.top(); s.pop();
            if (t->left) {
                TreeNode *r = t->left;
                while (r->right) r = r->right;
                r->right = t->right;
                t->right = t->left;
                t->left = NULL;
            }
            if (t->right) s.push(t->right);
        }
    }
};