树的操作-将树变为链表（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) {        //整体使用一个后续遍历        if(root==NULL ) return ;        flatten(root->left); //先处理左子树为链表,该语句返回时左子树已成链表        flatten(root->right);//再处理右子树为链表，该语句返回时右子树已成链表        //最后合并两个链表即可        TreeNode* tmp=root->right;//先备份一下右子树的根        root->right=root->left;//将左子树对应的链表链接在根的右子树上        TreeNode * tail=root;        for(;tail->right;tail=tail->right);//找到左子树链表的末尾        tail->right=tmp;//将其与右子树链表头链接        root->left=NULL;//最后不要忘了将左子树置空，如果没有这句，leetCode 将会提示“double free or corruption (fasttop): 0x0000000002785160 ***”的错误    }};