652. 寻找重复的子树 - 力扣（LeetCode)

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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];
};