【树】二叉树转为双向链表

目录：

1.【树】二叉树的非递归遍历（栈&Morris）

2.【树】二叉树转为双向链表

3.【树】树的广度遍历（1.打成一行 2.按层打成普通多行 3.按层打成之字形多行）

思路：

递归：

当前节点是root。

if（左子树不空）

找到root的前驱节点predecessor

把它跟root连起来

if（右子树不空）

找到root的后驱节点successor

把它跟root连起来。

最后返回的应为双向链表的头节点，所以再一直往左找就行啦。

**********************
二叉树->双向链表
**********************
Convert函数返回的是root节点，BinTreeToDLL函数返回的是最终答案（双向链表double-linked-list的头节点）

TreeNode* Convert(TreeNode* root)
{
    if (root == NULL)
        return root;
 
    // Convert the left subtree and link to root
    if (root->left != NULL) //此处先判断了左子树不空！ 所以while循环条件里无需担心predecessor为NULL的情况
    {
        // Convert the left subtree
        TreeNode* predecessor = Convert(root->left);
 
        // Find inorder predecessor. After this loop, predecessor will point to the inorder predecessor
        while(predecessor->right != NULL)
			predecessor = predecessor->right;
 
        // Make root as next of the predecessor and Make predecssor as previous of root
        predecessor->right = root;
        root->left = predecessor;
    }
 
    // Convert the right subtree and link to root
    if (root->right!=NULL)
    {
        // Convert the right subtree
        TreeNode* successor = Convert(root->right);
 
        // Find inorder successor. After this loop, successor will point to the inorder successor
        while(successor->left != NULL)
			successor = successor->left
		
        // Make root as previous of successor and Make successor as next of root
        successor->left = root;
        root->right = successor;
    }
 
    return root;
}
 

TreeNode* BinTreeToDLL(TreeNode* root) {
    if (root == NULL)
        return root;
 
    // Convert to DLL using Convert()
    TreeNode* listHead = Convert(root);
 
    // Convert() 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 (listHead->left != NULL)
        listHead = listHead->left;
 
    return listHead;
}

另一种AC写法：（没上面的好理解）

TreeNode* Convert(TreeNode* pRoot)
{
	//1.递归出口
	if(pRoot == NULL || pRoot->left == NULL && pRoot->right == NULL)
		return pRoot;
	//2.将左子树转成doubleList
	TreeNode* leftHead = Convert(pRoot->left);
	//3.找到左子树的尾结点
	TreeNode* leftTail = leftHead;
	while(leftTail != NULL && leftTail->right != NULL) {
		leftTail = leftTail->right;
	}
	//4.若左子树不为空，把pRoot加到【双向链接】doubleList尾部；否则，把leftHead指向pRoot【返回值】
	if(leftTail){
		leftTail->right = pRoot;
		pRoot->left = leftTail;
	}
	else
		leftHead = pRoot;
	//5.将右子树转成doubleList
	TreeNode* rightHead = Convert(pRoot->right);
	//6.若右子树不为空，就加到【双向链接】pRoot后面
	if(rightHead) {
		pRoot->right = rightHead;
		rightHead->left = pRoot;
	}
        
	return leftHead;
}