Question

1666. 改变二叉树的根节点

English Version

题目描述

给定一棵二叉树的根节点 root 和一个叶节点 leaf ，更改二叉树，使得 leaf 为新的根节点。

你可以按照下列步骤修改从 leaf 到 root 的路径中除 root 外的每个节点 cur ：

1. 如果 cur 有左子节点，则该子节点变为 cur 的右子节点。注意我们保证 cur 至多有一个子节点。
2. cur 的原父节点变为 cur 的左子节点。

返回修改后新树的根节点。

注意：确保你的答案在操作后正确地设定了 Node.parent （父节点）指针，否则会被判为错误答案。

 

示例 1:

输入: root = [3,5,1,6,2,0,8,null,null,7,4], leaf = 7
输出: [7,2,null,5,4,3,6,null,null,null,1,null,null,0,8]

示例 2:

输入: root = [3,5,1,6,2,0,8,null,null,7,4], leaf = 0
输出: [0,1,null,3,8,5,null,null,null,6,2,null,null,7,4]

 

提示:

• 树中节点的个数在范围 [2, 100] 内。
• -109 <= Node.val <= 109
• 所有的 Node.val 都是唯一的。
• leaf 存在于树中。

解法

方法一：自底向上模拟

从叶节点 leaf 开始，向上模拟翻转操作。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为二叉树节点个数。

"""
# Definition for a Node.
class Node:
  def __init__(self, val):
    self.val = val
    self.left = None
    self.right = None
    self.parent = None
"""


class Solution:
  def flipBinaryTree(self, root: "Node", leaf: "Node") -> "Node":
    cur = leaf
    p = cur.parent
    while cur != root:
      gp = p.parent
      if cur.left:
        cur.right = cur.left
      cur.left = p
      p.parent = cur
      if p.left == cur:
        p.left = None
      elif p.right == cur:
        p.right = None
      cur = p
      p = gp
    leaf.parent = None
    return leaf

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

class Solution {
  public Node flipBinaryTree(Node root, Node leaf) {
    Node cur = leaf;
    Node p = cur.parent;
    while (cur != root) {
      Node gp = p.parent;
      if (cur.left != null) {
        cur.right = cur.left;
      }
      cur.left = p;
      p.parent = cur;
      if (p.left == cur) {
        p.left = null;
      } else if (p.right == cur) {
        p.right = null;
      }
      cur = p;
      p = gp;
    }
    leaf.parent = null;
    return leaf;
  }
}

/*
// Definition for a Node->
class Node {
public:
  int val;
  Node* left;
  Node* right;
  Node* parent;
};
*/

class Solution {
public:
  Node* flipBinaryTree(Node* root, Node* leaf) {
    Node* cur = leaf;
    Node* p = cur->parent;
    while (cur != root) {
      Node* gp = p->parent;
      if (cur->left) {
        cur->right = cur->left;
      }
      cur->left = p;
      p->parent = cur;
      if (p->left == cur) {
        p->left = nullptr;
      } else if (p->right == cur) {
        p->right = nullptr;
      }
      cur = p;
      p = gp;
    }
    leaf->parent = nullptr;
    return leaf;
  }
};

/**
 * // Definition for a Node.
 * function Node(val) {
 *  this.val = val;
 *  this.left = null;
 *  this.right = null;
 *  this.parent = null;
 * };
 */

/**
 * @param {Node} node
 * @return {Node}
 */
var flipBinaryTree = function (root, leaf) {
  let cur = leaf;
  let p = cur.parent;
  while (cur != root) {
    const gp = p.parent;
    if (cur.left != null) {
      cur.right = cur.left;
    }
    cur.left = p;
    p.parent = cur;
    if (p.left == cur) {
      p.left = null;
    } else if (p.right == cur) {
      p.right = null;
    }
    cur = p;
    p = gp;
  }
  leaf.parent = null;
  return leaf;
};

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

public class Solution {
  public Node FlipBinaryTree(Node root, Node leaf) {
    Node cur = leaf;
    Node p = cur.parent;
    while (cur != root) {
      Node gp = p.parent;
      if (cur.left != null) {
        cur.right = cur.left;
      }
      cur.left = p;
      p.parent = cur;
      if (p.left == cur) {
        p.left = null;
      } else if (p.right == cur) {
        p.right = null;
      }
      cur = p;
      p = gp;
    }
    leaf.parent = null;
    return leaf;
  }
}