Question

17.12. BiNode

中文文档

Description

The data structure TreeNode is used for binary tree, but it can also used to represent a single linked list (where left is null, and right is the next node in the list). Implement a method to convert a binary search tree (implemented with TreeNode) into a single linked list. The values should be kept in order and the operation should be performed in place (that is, on the original data structure).

Return the head node of the linked list after converting.

Note: This problem is slightly different from the original one in the book.

 

Example:

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

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

Note:

• The number of nodes will not exceed 100000.

Solutions

Solution 1

# 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
}