面试题 36. 二叉搜索树与双向链表

"""
# 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 cur is None:
                return
            dfs(cur.left)
            if self.pre is None:
                self.head = cur
            else:
                self.pre.right = cur
            cur.left = self.pre
            self.pre = cur
            dfs(cur.right)
        if root is None:
            return None
        self.head = self.pre = None
        dfs(root)
        self.head.left = self.pre
        self.pre.right = self.head
        return self.head

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

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val,Node _left,Node _right) {
        val = _val;
        left = _left;
        right = _right;
    }
};
*/
class Solution {
    Node head;
    Node pre;
    public Node treeToDoublyList(Node root) {
        if (root == null) return null;
        dfs(root);
        head.left = pre;
        pre.right = head;
        return head;
    }

    private void dfs(Node cur) {
        if (cur == null) return;
        dfs(cur.left);
        if (pre == null) head = cur;
        else pre.right = cur;
        cur.left = pre;
        pre = cur;
        dfs(cur.right);
    }
}

/**
 * // Definition for a Node.
 * function Node(val,left,right) {
 *    this.val = val;
 *    this.left = left;
 *    this.right = right;
 * };
 */
/**
 * @param {Node} root
 * @return {Node}
 */
var treeToDoublyList = function (root) {
  function dfs(cur) {
    if (!cur) return;
    dfs(cur.left);
    if (!pre) head = cur;
    else pre.right = cur;
    cur.left = pre;
    pre = cur;
    dfs(cur.right);
  }
  if (!root) return null;
  let head, pre;
  dfs(root);
  head.left = pre;
  pre.right = head;
  return head;
};

class Solution {
public:
    Node* treeToDoublyList(Node* root) {
        if (root == NULL) return NULL;
        inorder(root);
        head->left = pre;
        pre->right = head;
        return head;
    }

private:
    Node *pre, *head;

    void inorder(Node* cur) {
        if (cur) {
            inorder(cur->left);
            if (pre)
                pre->right = cur;
            else
                head = cur;
            cur->left = pre;
            pre = cur;
            inorder(cur->right);
        }
    }
};