Question

652. 寻找重复的子树

English Version

题目描述

给你一棵二叉树的根节点 root ，返回所有 重复的子树 。

对于同一类的重复子树，你只需要返回其中任意 一棵 的根结点即可。

如果两棵树具有 相同的结构 和 相同的结点值 ，则认为二者是 重复 的。

 

示例 1：

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

示例 2：

输入：root = [2,1,1]
输出：[[1]]

示例 3：

输入：root = [2,2,2,3,null,3,null]
输出：[[2,3],[3]]

 

提示：

• 树中的结点数在 [1, 5000] 范围内。
• -200 <= Node.val <= 200

解法

方法一：后序遍历

后序遍历，序列化每个子树，用哈希表判断序列化的字符串出现次数是否等于 2，若是，说明这棵子树重复。

# 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 findDuplicateSubtrees(
    self, root: Optional[TreeNode]
  ) -> List[Optional[TreeNode]]:
    def dfs(root):
      if root is None:
        return '#'
      v = f'{root.val},{dfs(root.left)},{dfs(root.right)}'
      counter[v] += 1
      if counter[v] == 2:
        ans.append(root)
      return v

    ans = []
    counter = Counter()
    dfs(root)
    return ans

/**
 * 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 Map<String, Integer> counter;
  private List<TreeNode> ans;

  public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
    counter = new HashMap<>();
    ans = new ArrayList<>();
    dfs(root);
    return ans;
  }

  private String dfs(TreeNode root) {
    if (root == null) {
      return "#";
    }
    String v = root.val + "," + dfs(root.left) + "," + dfs(root.right);
    counter.put(v, counter.getOrDefault(v, 0) + 1);
    if (counter.get(v) == 2) {
      ans.add(root);
    }
    return v;
  }
}

/**
 * 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:
  unordered_map<string, int> counter;
  vector<TreeNode*> ans;

  vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
    dfs(root);
    return ans;
  }

  string dfs(TreeNode* root) {
    if (!root) return "#";
    string v = to_string(root->val) + "," + dfs(root->left) + "," + dfs(root->right);
    ++counter[v];
    if (counter[v] == 2) ans.push_back(root);
    return v;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func findDuplicateSubtrees(root *TreeNode) []*TreeNode {
  var ans []*TreeNode
  counter := make(map[string]int)
  var dfs func(root *TreeNode) string
  dfs = func(root *TreeNode) string {
    if root == nil {
      return "#"
    }
    v := strconv.Itoa(root.Val) + "," + dfs(root.Left) + "," + dfs(root.Right)
    counter[v]++
    if counter[v] == 2 {
      ans = append(ans, root)
    }
    return v
  }
  dfs(root)
  return ans
}

/**
 * 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 findDuplicateSubtrees(root: TreeNode | null): Array<TreeNode | null> {
  const map = new Map<string, number>();
  const res = [];
  const dfs = (root: TreeNode | null) => {
    if (root == null) {
      return '#';
    }
    const { val, left, right } = root;
    const s = `${val},${dfs(left)},${dfs(right)}`;
    map.set(s, (map.get(s) ?? 0) + 1);
    if (map.get(s) === 2) {
      res.push(root);
    }
    return s;
  };
  dfs(root);
  return res;
}

// 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;
use std::collections::HashMap;
impl Solution {
  fn dfs(
    root: &Option<Rc<RefCell<TreeNode>>>,
    map: &mut HashMap<String, i32>,
    res: &mut Vec<Option<Rc<RefCell<TreeNode>>>>
  ) -> String {
    if root.is_none() {
      return String::from('#');
    }

    let s = {
      let root = root.as_ref().unwrap().as_ref().borrow();
      format!(
        "{},{},{}",
        root.val.to_string(),
        Self::dfs(&root.left, map, res),
        Self::dfs(&root.right, map, res)
      )
    };
    *map.entry(s.clone()).or_insert(0) += 1;
    if *map.get(&s).unwrap() == 2 {
      res.push(root.clone());
    }
    return s;
  }

  pub fn find_duplicate_subtrees(
    root: Option<Rc<RefCell<TreeNode>>>
  ) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
    let mut map = HashMap::new();
    let mut res = Vec::new();
    Self::dfs(&root, &mut map, &mut res);
    res
  }
}