Question

面试题 17.12. BiNode

English Version

题目描述

二叉树数据结构TreeNode可用来表示单向链表（其中left置空，right为下一个链表节点）。实现一个方法，把二叉搜索树转换为单向链表，要求值的顺序保持不变，转换操作应是原址的，也就是在原始的二叉搜索树上直接修改。

返回转换后的单向链表的头节点。

注意：本题相对原题稍作改动

 

示例：

输入： [4,2,5,1,3,null,6,0]
输出： [0,null,1,null,2,null,3,null,4,null,5,null,6]

提示：

• 节点数量不会超过 100000。

解法

方法一：中序遍历

中序遍历过程中改变指针指向。

时间复杂度 $O(n)$。

同 897. 递增顺序查找树。

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, x):
#     self.val = x
#     self.left = None
#     self.right = None


class Solution:
  def convertBiNode(self, root: TreeNode) -> TreeNode:
    def dfs(root):
      if root is None:
        return
      nonlocal prev
      dfs(root.left)
      prev.right = root
      root.left = None
      prev = root
      dfs(root.right)

    dummy = TreeNode(val=0, right=root)
    prev = dummy
    dfs(root)
    return dummy.right

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode(int x) { val = x; }
 * }
 */
class Solution {
  private TreeNode prev;

  public TreeNode convertBiNode(TreeNode root) {
    TreeNode dummy = new TreeNode(0, null, root);
    prev = dummy;
    dfs(root);
    return dummy.right;
  }

  private void dfs(TreeNode root) {
    if (root == null) {
      return;
    }
    dfs(root.left);
    prev.right = root;
    root.left = null;
    prev = root;
    dfs(root.right);
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
  TreeNode* prev;

  TreeNode* convertBiNode(TreeNode* root) {
    TreeNode* dummy = new TreeNode(0, nullptr, root);
    prev = dummy;
    dfs(root);
    return dummy->right;
  }

  void dfs(TreeNode* root) {
    if (!root) return;
    dfs(root->left);
    prev->right = root;
    root->left = nullptr;
    prev = root;
    dfs(root->right);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func convertBiNode(root *TreeNode) *TreeNode {
  dummy := &TreeNode{Val: 0, Right: root}
  prev := dummy
  var dfs func(root *TreeNode)
  dfs = func(root *TreeNode) {
    if root == nil {
      return
    }
    dfs(root.Left)
    prev.Right = root
    root.Left = nil
    prev = root
    dfs(root.Right)
  }
  dfs(root)
  return dummy.Right
}