Question

897. 递增顺序搜索树

English Version

题目描述

给你一棵二叉搜索树的

 root ，请你 按中序遍历 将其重新排列为一棵递增顺序搜索树，使树中最左边的节点成为树的根节点，并且每个节点没有左子节点，只有一个右子节点。

 

示例 1：

输入：root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
输出：[1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

示例 2：

输入：root = [5,1,7]
输出：[1,null,5,null,7]

 

提示：

• 树中节点数的取值范围是 [1, 100]
• 0 <= Node.val <= 1000

解法

方法一：DFS 中序遍历

我们定义一个虚拟节点 $dummy$，初始时 $dummy$ 的右子节点指向根节点 $root$，定义一个指针 $prev$ 指向 $dummy$。

我们对二叉搜索树进行中序遍历，遍历过程中，每遍历到一个节点，就将 $prev$ 的右子节点指向它，然后将当前节点的左子节点置为空，再将当前节点赋值给 $prev$，以便于下一次遍历。

遍历结束后，原二叉搜索树被修改成只有右子节点的单链表，我们再将虚拟节点 $dummy$ 的右子节点返回即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉搜索树的节点个数。

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def increasingBST(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 = prev = TreeNode(right=root)
    dfs(root)
    return dummy.right

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  private TreeNode prev;
  public TreeNode increasingBST(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() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  TreeNode* increasingBST(TreeNode* root) {
    TreeNode* dummy = new TreeNode(0, nullptr, root);
    TreeNode* prev = dummy;
    function<void(TreeNode*)> dfs = [&](TreeNode* root) {
      if (!root) {
        return;
      }
      dfs(root->left);
      prev->right = root;
      root->left = nullptr;
      prev = root;
      dfs(root->right);
    };
    dfs(root);
    return dummy->right;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func increasingBST(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
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function increasingBST(root: TreeNode | null): TreeNode | null {
  const dummy = new TreeNode((right = root));
  let prev = dummy;
  const dfs = (root: TreeNode | null) => {
    if (!root) {
      return;
    }
    dfs(root.left);
    prev.right = root;
    root.left = null;
    prev = root;
    dfs(root.right);
  };
  dfs(root);
  return dummy.right;
}