652. 寻找重复的子树

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。

Python3JavaC++GoTypeScriptRust

# 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::cell::RefCell;
use std::collections::HashMap;
use std::rc::Rc;
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
    }
}

652. 寻找重复的子树

题目描述

解法

方法一：后序遍历

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