Question

面试题 04.10. 检查子树

English Version

题目描述

检查子树。你有两棵非常大的二叉树：T1，有几万个节点；T2，有几万个节点。设计一个算法，判断 T2 是否为 T1 的子树。

如果 T1 有这么一个节点 n，其子树与 T2 一模一样，则 T2 为 T1 的子树，也就是说，从节点 n 处把树砍断，得到的树与 T2 完全相同。

示例1:

 输入：t1 = [1, 2, 3], t2 = [2]
 输出：true

示例2:

 输入：t1 = [1, null, 2, 4], t2 = [3, 2]
 输出：false

提示：

1. 树的节点数目范围为[0, 20000]。

解法

方法一

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


class Solution:
  def checkSubTree(self, t1: TreeNode, t2: TreeNode) -> bool:
    def dfs(t1, t2):
      if t2 is None:
        return True
      if t1 is None:
        return False
      if t1.val == t2.val:
        return dfs(t1.left, t2.left) and dfs(t1.right, t2.right)
      return dfs(t1.left, t2) or dfs(t1.right, t2)

    return dfs(t1, t2)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode(int x) { val = x; }
 * }
 */
class Solution {
  public boolean checkSubTree(TreeNode t1, TreeNode t2) {
    if (t2 == null) {
      return true;
    }
    if (t1 == null) {
      return false;
    }
    if (t1.val == t2.val) {
      return checkSubTree(t1.left, t2.left) && checkSubTree(t1.right, t2.right);
    }
    return checkSubTree(t1.left, t2) || checkSubTree(t1.right, t2);
  }
}

/**
 * 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:
  bool checkSubTree(TreeNode* t1, TreeNode* t2) {
    if (!t2) return 1;
    if (!t1) return 0;
    if (t1->val == t2->val) return checkSubTree(t1->left, t2->left) && checkSubTree(t1->right, t2->right);
    return checkSubTree(t1->left, t2) || checkSubTree(t1->right, t2);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func checkSubTree(t1 *TreeNode, t2 *TreeNode) bool {
  if t2 == nil {
    return true
  }
  if t1 == nil {
    return false
  }
  if t1.Val == t2.Val {
    return checkSubTree(t1.Left, t2.Left) && checkSubTree(t1.Right, t2.Right)
  }
  return checkSubTree(t1.Left, t2) || checkSubTree(t1.Right, t2)
}

/**
 * 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 checkSubTree(t1: TreeNode | null, t2: TreeNode | null): boolean {
  if (t1 == null && t2 == null) {
    return true;
  }
  if (t1 == null || t2 == null) {
    return false;
  }
  if (t1.val === t2.val) {
    return checkSubTree(t1.left, t2.left) && checkSubTree(t1.right, t2.right);
  }
  return checkSubTree(t1.left, t2) || checkSubTree(t1.right, t2);
}

// 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 {
  fn dfs(t1: &Option<Rc<RefCell<TreeNode>>>, t2: &Option<Rc<RefCell<TreeNode>>>) -> bool {
    if t1.is_none() && t2.is_none() {
      return true;
    }
    if t1.is_none() || t2.is_none() {
      return false;
    }
    let r1 = t1.as_ref().unwrap().borrow();
    let r2 = t2.as_ref().unwrap().borrow();
    if r1.val == r2.val {
      return Self::dfs(&r1.left, &r2.left) && Self::dfs(&r1.right, &r2.right);
    }
    Self::dfs(&r1.left, t2) || Self::dfs(&r1.right, t2)
  }

  pub fn check_sub_tree(
    t1: Option<Rc<RefCell<TreeNode>>>,
    t2: Option<Rc<RefCell<TreeNode>>>
  ) -> bool {
    Self::dfs(&t1, &t2)
  }
}