【LeetCode笔记】114.Flatten Binary Tree to Linked List（有疑惑）

题目：

• Total Accepted: 119447
• Total Submissions: 348844
• Difficulty: Medium
• Contributor: LeetCode

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.

Hints:

If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

思路1：（求大神们指正）

原始的树前序遍历，存入队列，然后顺序读出，一直往树的右子树放置。这样的思路我把答案输出了，我觉得是对的呀，不知道为啥leetcode一直判断我是错的，如果有大神能来指正，真心感谢！

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:    void p(TreeNode* root,queue<int> &v){        if (root ==NULL)            return;        else{            v.push(root->val);            if(root->left!=NULL)                p(root->left,v);            if(root->right!=NULL)                p(root->right,v);        }    }    void flatten(TreeNode* root) {        if(root==NULL)            return;        else{            queue<int> v;            p(root,v);            TreeNode* t = new TreeNode(v.front());            TreeNode* temp = t;            v.pop();                    while(!v.empty()){                TreeNode* l = new TreeNode(v.front());                t->right = l;                t = t->right;                v.pop();            }            root = temp;

//这一段printf是为了验证我的答案到底都不对才写的，就是把root节点打印出来看看到底是什么，结果显示是对的呀。。。            if(root!=NULL)                printf("root = %d\n",root->val);            else                printf("root null!\n");            if(root->left!=NULL)                printf("ok2  %d\n",root->left->val);            else                printf("root->left null\n");            if(root->right!=NULL)                printf("root->right =   %d\n",root->right->val);            else                printf("root->right null\n");                }    }};

结果：

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Submission Result: Wrong Answer More Details 

Input:[1,2]

Output:[1,2]

Expected:[1,null,2]

Stdout:root = 1root->left nullroot->right = 2

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小白不明白的是，为啥我输出的答案和标准答案是一样的，但是系统判错？

更新思路：把左子树的所有节点按照存入右子树里，然后把原来右子树的节点存入左子树最右的节点中。递归撸平右子树。

/** * 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;        if(root->left!=NULL){            TreeNode* temp = root->left;            while(temp->right!=NULL)                temp = temp->right;            temp->right = root->right;            root->right = root->left;            root->left = NULL;                    }        flatten(root->right);    }};