樹的操作-將樹變爲鏈表（Flatten Binary Tree to Linked List）

1 Orginal problem:

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

2 Solution

using post-order traverse of tree, you can solve it.
The code in C++ as follow:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void flatten(TreeNode* root) {
        //整體使用一個後續遍歷

        if(root==NULL ) return ;

        flatten(root->left); //先處理左子樹爲鏈表,該語句返回時左子樹已成鏈表
        flatten(root->right);//再處理右子樹爲鏈表，該語句返回時右子樹已成鏈表

        //最後合併兩個鏈表即可
        TreeNode* tmp=root->right;//先備份一下右子樹的根
        root->right=root->left;//將左子樹對應的鏈表鏈接在根的右子樹上

        TreeNode * tail=root;
        for(;tail->right;tail=tail->right);//找到左子樹鏈表的末尾
        tail->right=tmp;//將其與右子樹鏈表頭鏈接
        root->left=NULL;//最後不要忘了將左子樹置空，如果沒有這句，leetCode 將會提示“double free or corruption (fasttop): 0x0000000002785160 ***”的錯誤
    }
};