[LeetCode] Flatten Binary Tree to Linked List 解题报告

最新推荐文章于 2022-02-22 10:11:49 发布

原创最新推荐文章于 2022-02-22 10:11:49 发布 · 280 阅读

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本文介绍如何使用递归方法将二叉树扁平化为链表，通过转换节点链接实现从根节点到叶子节点的顺序连接。

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

The flattened tree should look like:

click to show hints.

» Solve this problem

[解题思路]
递归解法。对于任一节点，flatten左树，然后节点插入左树最左边，成为新的头节点。flatten右树，右树最左边接上新链表的最右节点。

[Code]

1:  void flatten(TreeNode *root) {  
2:       // Start typing your C/C++ solution below  
3:       // DO NOT write int main() function  
4:       if(root == NULL)  
5:            return;  
6:       ConvertToLink(root);  
7:  }  
8:  TreeNode* ConvertToLink(TreeNode* node)  
9:  {  
10:       if(node->left == NULL && node->right == NULL)  
11:            return node;  
12:       TreeNode* rHead = NULL;  
13:       if(node->right != NULL)  
14:           rHead = ConvertToLink(node->right);               
15:       TreeNode* p = node;  
16:       if(node->left!=NULL)  
17:       {  
18:            TreeNode* lHead = ConvertToLink(node->left);   
19:            node->right = lHead;  
20:            lHead->left = NULL;  
21:            node->left = NULL;  
22:            while(p->right!=NULL)  
23:                 p = p->right;  
24:       }       
25:       if(rHead != NULL)  
26:       {  
27:            p->right = rHead;  
28:            rHead->left = NULL;  
29:       }  
30:       return node;  
31:  }

[已犯错误]
1. Line 13~14
刚开始的时候，首先flatten左树，然后处理右树。但是这样会导致处理右树的时候，节点的值已经在处理树的时候被破坏了。比如树为{1,2,3}，
1
/
2 3
如果先处理左树的话，当执行node->right = lhead的时候，右节点就已经被破坏了，node->right指向了2，而不是3。
1

2 (3)
当然，也可以用一个变量在处理左树前，保存右树地址。但是没必要，先处理右树就好了。
2. Line 22~23
该循环是用于将p指针遍历到左树链表的最右节点。第一版时，这个循环是放在if语句以外，这就导致了，不必要的迭代了。比如当输入为{1,#,2}时，这个while循环会导致p指针遍历到右子树的最右节点，这显然是错的。
3. Line 20, 28
不要忘了清空每一个指针，在新的链表中，左指针没必要保留。

Update 08/25/2014 being asked this question today. But the interviewer asked for an in-order flatten.
Review previous solution. Actually, I made it too complicate. If travel this tree in pre-order, from the hint, it is easy to construct the linked list.

1:       void flatten(TreeNode *root) {  
2:            if(root == NULL) return;  
3:            TreeNode* right = root->right;  
4:            if(lastVisitedNode != NULL)  
5:            {  
6:                 lastVisitedNode->left = NULL;  
7:                 lastVisitedNode->right = root;  
8:            }  
9:            lastVisitedNode = root;  
10:            flatten(root->left);  
11:            flatten(right);  
12:       }

pre-order is simple because the root always is the head of flatten list. But if flatten the tree with in-order sequence, need extra parameter to track the head and tail of each flattened sun-tree.
For example, below binary tree.

If we flatten it with in-order, the process should like below. And here I use the left pointer of head node to track the tail node.

1:  TreeNode* flatten(TreeNode *root) {  
2:       if (root == NULL) return NULL;  
3:       TreeNode* rightTree = root->right;  
4:       TreeNode* newHead = root;  
5:       TreeNode* leftList = flatten(root->left);  
6:       if (leftList != NULL)  
7:       {  
8:            newHead = leftList;  
9:            TreeNode* tail = leftList->left;  
10:            tail->right = root;  
11:            root->left = tail;  
12:            leftList->left = root;  
13:       }  
14:       TreeNode* rightList = flatten(rightTree);  
15:       if (rightList != NULL)  
16:       {  
17:            root->right = rightList;  
18:            newHead->left = rightList->left;  
19:            rightList->left = root;  
20:       }  
21:       return newHead;  
22:  }