Flatten Binary Tree to Linked List

Problem Description:

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.

思路一，按照题意从根节点开始，如果当前节点有左孩子，就将它的左孩子添加到自己和右孩子之间，这里每次需要找到左孩子最右边的节点，连接到当前右孩子。然后依次往右处理自己的右孩子，直到右孩子为空。具体代码如下：

/** * Definition for binary tree * 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;        while (root != NULL)         {if (root->left != NULL) {TreeNode *p = root->left;while (p->right != NULL) <span style="font-family: Arial, Helvetica, sans-serif;">//找到左孩子的最右边节点</span>{p = p->right;}p->right = root->right;root->right = root->left;root->left = NULL;}root = root->right;}    }};

思路二：可以看出变换后的树实际上是按照先序遍历的顺序排列的，因此可以利用中序递归遍历，记录先序遍历的前驱结点，依次调整每个孩子的左右子树，这里需要注意的是遍历过程中需要先将当前节点的右子树记录下来，再调整当前节点的左右子树，然后递归调整左右子树。具体代码如下：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode *pre=NULL;    void flatten(TreeNode *root) {        if(root==NULL)            return;        TreeNode *lastright=root->right;//记录当前节点的右子树        if(pre)        {            pre->left=NULL;            pre->right=root;        }        pre=root;        flatten(root->left);        flatten(lastright);    }};