剑指offer66题--Java实现，c++实现和python实现 26.二叉搜索树与双向链表

题目描述

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。

C++

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == nullptr) return nullptr;
        TreeNode* pre = nullptr;
         
        convertHelper(pRootOfTree, pre);
         
        TreeNode* res = pRootOfTree;
        while(res ->left)
            res = res ->left;
        return res;
    }
     
    void convertHelper(TreeNode* cur, TreeNode*& pre)
    {
        if(cur == nullptr) return;
         
        convertHelper(cur ->left, pre);
         
        cur ->left = pre;
        if(pre) pre ->right = cur;
        pre = cur;
         
        convertHelper(cur ->right, pre);
         
         
         
    }
};

java 

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    private TreeNode leftHead = null;
    private TreeNode rightHead = null;

    public TreeNode Convert(TreeNode pRootOfTree) {
        if (pRootOfTree == null) return  null;

        Convert(pRootOfTree.left);

        if (leftHead == null) {
            leftHead = pRootOfTree;
            rightHead = pRootOfTree;
        }
        else {
            rightHead.right = pRootOfTree;
            pRootOfTree.left = rightHead;
            rightHead = pRootOfTree;
        }

        Convert(pRootOfTree.right);

        return leftHead;
    }
}

python

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def Convert(self,pRootOfTree):
        if pRootOfTree == None:
            return None
        if not pRootOfTree.left and not pRootOfTree.right:
            return pRootOfTree
        left = pRootOfTree.left
        if left:
            self.Convert(left)
            while left.right:
                left = left.right
            pRootOfTree.left,left.right = left,pRootOfTree
            
        right = pRootOfTree.right
        if right:
            self.Convert(right)
            while right.left:
                right = right.left
            pRootOfTree.right, right.left = right,pRootOfTree
            
        while pRootOfTree.left:
            pRootOfTree = pRootOfTree.left
        return pRootOfTree