Question

114. 二叉树展开为链表

English Version

题目描述

给你二叉树的根结点 root ，请你将它展开为一个单链表：

• 展开后的单链表应该同样使用 TreeNode ，其中 right 子指针指向链表中下一个结点，而左子指针始终为 null 。
• 展开后的单链表应该与二叉树 先序遍历 顺序相同。

 

示例 1：

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

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [0]
输出：[0]

 

提示：

• 树中结点数在范围 [0, 2000] 内
• -100 <= Node.val <= 100

 

进阶：你可以使用原地算法（O(1) 额外空间）展开这棵树吗？

解法

方法一：寻找前驱节点

先序遍历的访问顺序是“根、左子树、右子树”，左子树最后一个节点访问完后，接着会访问根节点的右子树节点。

因此，对于当前节点，如果其左子节点不为空，我们找到左子树的最右节点，作为前驱节点，然后将当前节点的右子节点赋给前驱节点的右子节点。然后将当前节点的左子节点赋给当前节点的右子节点，并将当前节点的左子节点置为空。然后将当前节点的右子节点作为下一个节点，继续处理，直至所有节点处理完毕。

时间复杂度 $O(n)$，其中 $n$ 是树中节点的个数。空间复杂度 $O(1)$。

# 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 flatten(self, root: Optional[TreeNode]) -> None:
    """
    Do not return anything, modify root in-place instead.
    """
    while root:
      if root.left:
        pre = root.left
        while pre.right:
          pre = pre.right
        pre.right = root.right
        root.right = root.left
        root.left = None
      root = root.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 {
  public void flatten(TreeNode root) {
    while (root != null) {
      if (root.left != null) {
        // 找到当前节点左子树的最右节点
        TreeNode pre = root.left;
        while (pre.right != null) {
          pre = pre.right;
        }

        // 将左子树的最右节点指向原来的右子树
        pre.right = root.right;

        // 将当前节点指向左子树
        root.right = root.left;
        root.left = null;
      }
      root = 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:
  void flatten(TreeNode* root) {
    while (root) {
      if (root->left) {
        TreeNode* pre = root->left;
        while (pre->right) {
          pre = pre->right;
        }
        pre->right = root->right;
        root->right = root->left;
        root->left = nullptr;
      }
      root = root->right;
    }
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func flatten(root *TreeNode) {
  for root != nil {
    if root.Left != nil {
      pre := root.Left
      for pre.Right != nil {
        pre = pre.Right
      }
      pre.Right = root.Right
      root.Right = root.Left
      root.Left = nil
    }
    root = root.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)
 *   }
 * }
 */

/**
 Do not return anything, modify root in-place instead.
 */
function flatten(root: TreeNode | null): void {
  while (root !== null) {
    if (root.left !== null) {
      let pre = root.left;
      while (pre.right !== null) {
        pre = pre.right;
      }
      pre.right = root.right;
      root.right = root.left;
      root.left = null;
    }
    root = root.right;
  }
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  #[allow(dead_code)]
  pub fn flatten(root: &mut Option<Rc<RefCell<TreeNode>>>) {
    if root.is_none() {
      return;
    }
    let mut v: Vec<Option<Rc<RefCell<TreeNode>>>> = Vec::new();
    // Initialize the vector
    Self::pre_order_traverse(&mut v, root);
    // Traverse the vector
    let n = v.len();
    for i in 0..n - 1 {
      v[i].as_ref().unwrap().borrow_mut().left = None;
      v[i].as_ref().unwrap().borrow_mut().right = v[i + 1].clone();
    }
  }

  #[allow(dead_code)]
  fn pre_order_traverse(
    v: &mut Vec<Option<Rc<RefCell<TreeNode>>>>,
    root: &Option<Rc<RefCell<TreeNode>>>
  ) {
    if root.is_none() {
      return;
    }
    v.push(root.clone());
    let left = root.as_ref().unwrap().borrow().left.clone();
    let right = root.as_ref().unwrap().borrow().right.clone();
    Self::pre_order_traverse(v, &left);
    Self::pre_order_traverse(v, &right);
  }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {void} Do not return anything, modify root in-place instead.
 */
var flatten = function (root) {
  while (root) {
    if (root.left) {
      let pre = root.left;
      while (pre.right) {
        pre = pre.right;
      }
      pre.right = root.right;
      root.right = root.left;
      root.left = null;
    }
    root = root.right;
  }
};

方法二

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func flatten(root *TreeNode) {
  for root != nil {
    left, right := root.Left, root.Right
    root.Left = nil
    if left != nil {
      root.Right = left
      for left.Right != nil {
        left = left.Right
      }
      left.Right = right
    }
    root = root.Right
  }
}