leetcode - Flatten Binary Tree to Linked List

题目：Flatten Binary Tree to Linked List

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.

 

个人思路：

1、创建一个队列，把节点按照先序遍历的顺序入队

2、把节点挨个出队，前一个出队节点是后一个出队节点的父节点，后一个出队节点是前一个出队节点的右孩子节点，且每个非叶节点的左孩子为空

 

代码：

 1 #include <stddef.h>
 2 #include <iostream>
 3 #include <queue>
 4 
 5 using namespace std;
 6 
 7 struct TreeNode
 8 {
 9     int val;
10     TreeNode *left;
11     TreeNode *right;
12     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
13 };
14 
15 class Solution
16 {
17 public:
18     void flatten(TreeNode *root)
19     {
20         if (!root)
21         {
22             return;
23         }
24 
25         queue<TreeNode *> preNodes;
26         preorder(root, preNodes);
27         root = preNodes.front();
28         preNodes.pop();
29         TreeNode * currentNode = root;
30         while (!preNodes.empty())
31         {
32             TreeNode *tmp = preNodes.front();
33             preNodes.pop();
34             currentNode->right = tmp;
35             currentNode->left = NULL;
36             currentNode = currentNode->right;
37         }
38     }
39 private:
40     void preorder(TreeNode *root, queue<TreeNode *> &preNodes)
41     {
42         if (!root)
43         {
44             return;
45         }
46         preNodes.push(root);
47         preorder(root->left, preNodes);
48         preorder(root->right, preNodes);
49     }
50 };
View Code

 

在网上看到一个比较普遍的递归算法，链接：http://blog.unieagle.net/2012/12/16/leetcode-problemflatten-binary-tree-to-linked-list/，里面有比较详细的思路解析，可以参考，大致思路是这样：要扁平化一棵树，则扁平化根节点的左子树和右子树，然后将左子树放到根节点的右孩子位置，右子树放到左子树的叶节点的右孩子位置

 

代码：

 1 #include <stddef.h>
 2 #include <iostream>
 3 #include <queue>
 4 
 5 using namespace std;
 6 
 7 struct TreeNode
 8 {
 9     int val;
10     TreeNode *left;
11     TreeNode *right;
12     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
13 };
14 
15 class Solution
16 {
17 public:
18     void flatten(TreeNode *root)
19     {
20         /*
21         if (!root)
22         {
23             return;
24         }
25 
26         queue<TreeNode *> preNodes;
27         preorder(root, preNodes);
28         root = preNodes.front();
29         preNodes.pop();
30         TreeNode * currentNode = root;
31         while (!preNodes.empty())
32         {
33             TreeNode *tmp = preNodes.front();
34             preNodes.pop();
35             currentNode->right = tmp;
36             currentNode->left = NULL;
37             currentNode = currentNode->right;
38         }
39         */
40 
41         if (!root)
42         {
43             return;
44         }
45         if (root->left)
46         {
47             flatten(root->left);
48         }
49         if (root->right)
50         {
51             flatten(root->right);
52         }
53         if (!root->left)
54         {
55             return;
56         }
57 
58         TreeNode *left_leaf = root->left;
59         while (left_leaf->right)
60         {
61             left_leaf = left_leaf->right;
62         }
63         left_leaf->right = root->right;
64         root->right = root->left;
65         root->left = NULL;
66     }
67 private:
68     void preorder(TreeNode *root, queue<TreeNode *> &preNodes)
69     {
70         if (!root)
71         {
72             return;
73         }
74         preNodes.push(root);
75         preorder(root->left, preNodes);
76         preorder(root->right, preNodes);
77     }
78 };
View Code

 

链接中还有一个非递归的思路，大家也可以看看

posted on 2014-06-20 14:01  laihaiteng  阅读(204)  评论(0)  编辑  收藏  举报

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