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Friday, December 21, 2012

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


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
» 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:  }  


















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