[LeetCode 题解]: 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

Hints:

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

题意：将任意一颗二叉树转换成一颗仅有左子树的二叉树。

要求：（1）不得使用额外存储空间，就地排序整理。

（2）修改之后的树为仅有左子树的二叉树，节点顺序为原二叉树的先序遍历。

思路：说是结果为先序遍历，但是递归方式为后序遍历方式。

class Solution {
public:
    void flatten(TreeNode *root) {
        if(root==NULL) return;
        flatten(root->left);
        flatten(root->right);
        if(root->left==NULL) return;
        TreeNode *p = root->left;
        while(p->right) p=p->right;
        p->right=root->right;
        root->right=root->left;
        root->left=NULL;
    }
};

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[LeetCode 题解]: Flatten Binary Tree to Linked List