leetcode 114. Flatten Binary Tree to Linked List 扁平化二叉树+前序遍历

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
click to show hints.

这道题很简答，就是做一次中序遍历，然后把它做回扁平化就可以了。

代码如下;

import java.util.ArrayList;import java.util.List;/*class TreeNode{      int val;      TreeNode left;      TreeNode right;      TreeNode(int x) { val = x; }}*/public class Solution {    List<TreeNode> res=new ArrayList<TreeNode>();    public void flatten(TreeNode root)    {        if(root==null)            return ;        preOrder(root);        for(int i=0;i<res.size()-1;i++)        {            TreeNode one=res.get(i);            TreeNode two=res.get(i+1);            one.left=two.left=null;            one.right=two;        }        TreeNode last=res.get(res.size()-1);        last.left=last.right=null;    }    void preOrder(TreeNode root)     {        if(root!=null)        {            res.add(root);            preOrder(root.left);            preOrder(root.right);        }    }}

下面是C++的做法，就是先做一个二叉树的前序遍历，然后按照要求去做就可以了

代码如下：

#include <iostream>#include <vector>using namespace std;/*struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {}};*/class Solution {public:    vector<TreeNode*> one;    void flatten(TreeNode* root)     {        getAll(root);        if (one.size() >= 2)        {            for (int i = 1; i < one.size(); i++)            {                one[i - 1]->right = one[i];                one[i - 1]->left = NULL;            }            one[one.size() - 1]->left = NULL;            one[one.size() - 1]->right = NULL;        }        else            return;    }    void getAll(TreeNode* root)    {        if (root == NULL)            return;        else        {            one.push_back(root);            getAll(root->left);            getAll(root->right);        }    }};