数据结构与算法之LeetCode-652. 寻找重复的子树

news/2024/7/7 14:56:10

652. 寻找重复的子树 - 力扣（LeetCode）

/**
 * 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 {TreeNode[]}
 */
var findDuplicateSubtrees = function(root) {
    const seen = new Map();
    const repeat = new Set();
    const dfs = (node) => {
        if (!node) {
            return "";
        }
        let sb = '';
        sb += node.val;
        sb += "(";
        sb += dfs(node.left);
        sb += ")(";
        sb += dfs(node.right);
        sb += ")";
        if (seen.has(sb)) {
            repeat.add(seen.get(sb));
        } else {
            seen.set(sb, node);
        }
        return sb;
    }
    dfs(root);
    return [...repeat];
};

执行结果：通过

执行用时：108 ms, 在所有 JavaScript 提交中击败了67.16%的用户

内存消耗：48.7 MB, 在所有 JavaScript 提交中击败了43.14%的用户

通过测试用例：176 / 176

var findDuplicateSubtrees = function(root) {
    const seen = new Map();
    const repeat = new Set();
    let idx = 0;
    const dfs = (node) => {
        if (!node) {
            return 0;
        }
        const tri = [node.val, dfs(node.left), dfs(node.right)];
        const hash = tri.toString();
        if (seen.has(hash)) {
            const pair = seen.get(hash);
            repeat.add(pair[0]);
            return pair[1];
        } else {
            seen.set(hash, [node, ++idx]);
            return idx;
        }
    }
    dfs(root);
    return [...repeat];
};