54. Flatten Binary Tree to Linked List

1. Flatten Binary Tree to Linked List My Submissions QuestionEditorial Solution
   
   Total Accepted: 81373 Total Submissions: 261933 Difficulty: Medium
   
   Given a binary tree, flatten it to a linked list in-place.

For example,

Given

     1
    / \
   2   5
  / \   \
 3   4   6

The flattened tree should look like:

1

\

2

\

3

\

4

\

5

\

6

思路：

用vector按照先序遍历的顺序把指针全部存下来，然后依次用右指针链接起来

时间复杂度：O(n)

空间复杂度：O(n)

/**
 * 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:
    void flatten(TreeNode* root) {
        if(root==NULL)return;
        stack<TreeNode*> stn;  //用栈来进行先序遍历，注意这里不允许递归，除非另写函数
        vector<TreeNode*> vtn;
        stn.push(root);
        while(!stn.empty()){
            TreeNode *p = stn.top();
            vtn.push_back(p);
            stn.pop();
            if(p->right!=NULL)stn.push(p->right);
            if(p->left!=NULL)stn.push(p->left);
        }
        for(int i=0;i<vtn.size()-1;++i){
            vtn[i]->right = vtn[i+1];
            vtn[i]->left = NULL;
        }
        vtn[vtn.size()-1]->left = NULL;vtn[vtn.size()-1]->right = NULL;
    }
};