剑指offer | 二叉搜索树与双向链表 | 31

思路分析

二叉搜索树和双向链表其实是以中序遍历的方式来建立一个双向链表.
中序遍历

# 打印中序遍历
def dfs(root):
    if not root: return
    dfs(root.left)  # 左
    print(root.val) # 根
    dfs(root.right) # 右

主要有3个指针

pre,head,cur

cpp

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;

    Node() {}

    Node(int _val) {
        val = _val;
        left = NULL;
        right = NULL;
    }

    Node(int _val, Node* _left, Node* _right) {
        val = _val;
        left = _left;
        right = _right;
    }
};
*/
class Solution {
public:
    Node *head=NULL,*pre=NULL;
    
    void dfs(Node *cur){
        if(!cur)return;
        dfs(cur->left);
        
        if(!pre)head=cur;
        else pre->right=cur;
        
        cur->left = pre;
        pre = cur;
        dfs(cur->right);
    }

    Node* treeToDoublyList(Node* root) {
        if(!root)return NULL;
        dfs(root);
        pre->right=head,head->left=pre;
        return head;
    }
};

python

"""
# Definition for a Node.
class Node:
    def __init__(self, val, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
"""
class Solution:
    def treeToDoublyList(self, root: ‘Node‘) -> ‘Node‘:
        def dfs(cur):
            if not cur: return
            dfs(cur.left) # 递归左子树
            if self.pre: # 修改节点引用
                self.pre.right, cur.left = cur, self.pre
            else: # 记录头节点
                self.head = cur
            self.pre = cur # 保存 cur
            dfs(cur.right) # 递归右子树
        
        if not root: return
        self.pre = None
        dfs(root)
        self.head.left, self.pre.right = self.pre, self.head
        return self.head

剑指offer | 二叉搜索树与双向链表 | 31

原文：https://www.cnblogs.com/Rowry/p/14365328.html