二叉树转双向链表

http://www.cppblog.com/whspecial/archive/2014/01/03/205123.html

将排序二叉树转换成双向链表
Posted on 2014-01-03 00:41 whspecial 阅读(2169) 评论(0) 编辑 收藏 引用 所属分类: 算法&&数据结构
将排序二叉树转化成双向链表，应该是一道很常见的面试题目，网上的实现比较多，有用递归也有用中序遍历法的。看到一位外国友人的实现，还是比较清晰的，思路如下：
1，如果左子树不为null，处理左子树

1.a）递归转化左子树为双向链表；

1.b）找出根结点的前驱节点（是左子树的最右的节点）

1.c）将上一步找出的节点和根结点连接起来

2，如果右子树不为null，处理右子树（和上面的很类似）

1.a）递归转化右子树为双向链表；

1.b）找出根结点的后继节点（是右子树的最左的节点）

1.c）将上一步找出的节点和根结点连接起来

3，找到最左边的节点并返回

附上国外友人的链接：http://www.geeksforgeeks.org/in-place-convert-a-given-binary-tree-to-doubly-linked-list/

下面是代码实现：

bintree2listUtil函数返回的node＊ 是root节点，bintree2list函数返回的是头节点

 This is the core function to convert Tree to list. This function follows  steps 1 and 2 of the above algorithm */node* bintree2listUtil(node* root){    // Base case    if (root == NULL)        return root;    // Convert the left subtree and link to root    if (root->left != NULL)    {        // Convert the left subtree        node* left = bintree2listUtil(root->left);        // Find inorder predecessor. After this loop, left        // will point to the inorder predecessor        for (; left->right!=NULL; left=left->right);        // Make root as next of the predecessor        left->right = root;        // Make predecssor as previous of root        root->left = left;    }    // Convert the right subtree and link to root    if (root->right!=NULL)    {        // Convert the right subtree        node* right = bintree2listUtil(root->right);        // Find inorder successor. After this loop, right        // will point to the inorder successor        for (; right->left!=NULL; right = right->left);        // Make root as previous of successor        right->left = root;        // Make successor as next of root        root->right = right;    }    return root;}// The main function that first calls bintree2listUtil(), then follows step 3 //  of the above algorithmnode* bintree2list(node *root){    // Base case    if (root == NULL)        return root;    // Convert to DLL using bintree2listUtil()    root = bintree2listUtil(root);    // bintree2listUtil() returns root node of the converted    // DLL.  We need pointer to the leftmost node which is    // head of the constructed DLL, so move to the leftmost node    while (root->left != NULL)        root = root->left;    return (root);