剑指Offer-26.二叉搜索树与双向链表(C++/Java)

题目：

输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向。

分析：

创建两个指针，分别指向要处理的当前元素和当前元素的前一个元素。利用中序遍历找到二叉树中最小的元素，实际上也是双向链表的最左端的值。依次将当前元素和前一个元素用left和right连接起来，在处理右子树之前，要将前一个元素改成当前的元素，以便后续元素的处理。如图：

图出自：https://www.jianshu.com/p/7607d8519c59

 

 

 当然还可以根据中序遍历，将所有节点存储起来，然后再重新按顺序连接所有节点，因为中序遍历的节点已经有序，直接前后相连即可。

程序：

C++

class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == nullptr)
            return nullptr;
        if(pRootOfTree->left == nullptr && pRootOfTree->right == nullptr)
            return pRootOfTree;
        helper(pRootOfTree);
        for(int i = 0; i < node.size()-1; ++i){
            node[i]->right = node[i+1];
            node[i+1]->left = node[i];
        }
        return node[0];
    }
    void helper(TreeNode* root){
        if(root == nullptr)
            return;
        helper(root->left);
        node.push_back(root);
        helper(root->right);
    }
private:
    vector<TreeNode*> node;
};

class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == nullptr)
            return nullptr;
        if(pRootOfTree->left == nullptr && pRootOfTree->right == nullptr)
            return pRootOfTree;
        TreeNode* pre = nullptr;
        helper(pRootOfTree, pre);
        while(pRootOfTree->left != nullptr){
            pRootOfTree = pRootOfTree->left;
        }
        return pRootOfTree;
    }
    void helper(TreeNode* cur, TreeNode* &pre){
        if(cur == nullptr)
            return;
        if(cur->left != nullptr)
            helper(cur->left, pre);
        cur->left = pre;
        if(pre != nullptr)
            pre->right = cur;
        pre = cur;
        if(cur->right != nullptr){
            helper(cur->right, pre);
        }
    }
};

Java

public class Solution {
    public TreeNode Convert(TreeNode pRootOfTree) {
        if(pRootOfTree == null)
            return null;
        if(pRootOfTree.left == null && pRootOfTree.right == null)
            return pRootOfTree;
        //TreeNode pre = null;
        helper(pRootOfTree);
        while(pRootOfTree.left != null){
            pRootOfTree = pRootOfTree.left;
        }
        return pRootOfTree;
    }
    public void helper(TreeNode cur){
        if(cur == null)
            return;
        if(cur.left != null)
            helper(cur.left);
        cur.left = pre;
        if(pre != null)
            pre.right = cur;
        pre = cur;
        if(cur.right != null){
            helper(cur.right);
        }
    }
    private TreeNode pre = null;
}