508. Subtree elements with the most occurrences and | Algorithm (leetcode, with mind map + all solutions) 300 questions

Zero title: Algorithm (leetcode, with mind map + all solutions) of 300 questions (508) with the most frequent subtree elements and

a topic description

Two solutions overview (mind map)

All three solutions

1 Scenario 1

1) Code:

 // 方案1 “自己。后续遍历法”。
// 用时：14分钟。

// 思路：
// 1）状态初始化：resMap = new Map() 。
// 2）调用递归函数 updateRootValByDfs(root) 。
// 3）求得 出现次数最多的子树元素和 resMaxCount = getMaxCountByMap(resMap) 。
// 4）若 当前值（即 key ）出现的次数 val 与 resMaxCount，
// 则 将 val 放入 resList 中。
// 5）返回结果 resList 。
var findFrequentTreeSum = function(root) {
    const updateRootValByDfs = (curRoot = null) => {
        // 1）递归出口
        if (!curRoot) {
            return;
        }
        const {left, right} = curRoot;
        if (!left && !right) {
            if (resMap.has(curRoot.val)) {
                resMap.set(curRoot.val, resMap.get(curRoot.val) + 1);
            }
            else {
                resMap.set(curRoot.val, 1);
            }
            return;
        }

        // 2）递归主体
        // 2.1）先更新 left 节点上的 val 。
        updateRootValByDfs(left);
        // 2.2）接着更新 right 节点上的 val 。
        updateRootValByDfs(right);
        // 2.3）最后更新 curRoot 节点上的 val 。
        if (left) {
            curRoot.val += (left.val);
        }
        if (right) {
            curRoot.val += (right.val);
        }

        if (resMap.has(curRoot.val)) {
            resMap.set(curRoot.val, resMap.get(curRoot.val) + 1);
        }
        else {
            resMap.set(curRoot.val, 1);
        }
    };
    const getMaxCountByMap = (map = new Map()) => {
        let resMaxCount = Number.NEGATIVE_INFINITY;

        for (const [key, val] of map) {
            resMaxCount = Math.max(resMaxCount, val);
        }

        return resMaxCount;
    };

    // 边界（根据提示，如下代码可忽略）
    if (!root) {
        return [];
    }

    // 1）状态初始化：resMap = new Map() 。
    let resMap = new Map();

    // 2）调用递归函数 updateRootValByDfs(root) 。
    updateRootValByDfs(root);

    // 3）求得 出现次数最多的子树元素和 resMaxCount = getMaxCountByMap(resMap) 。
    let resMaxCount = getMaxCountByMap(resMap),
        resList = [];
    
    // 4）若 当前值（即 key ）出现的次数 val 与 resMaxCount，
    // 则 将 val 放入 resList 中。
    for (const [key, val] of resMap) {
        if (val === resMaxCount) {
            resList.push(key);
        }
    }

    // 5）返回结果 resList 。
    return resList;
};

2 Option 2

1) Code:

 // 方案2 “官方。深度优先遍历法（本质：跟自己的方案1大体上是一样的）”。
// 参考：
// 1）https://leetcode.cn/problems/most-frequent-subtree-sum/solution/chu-xian-ci-shu-zui-duo-de-zi-shu-yuan-s-kdjc/

// 思路：
// 1）状态初始化：esMap = new Map(), resMaxCount = Number.NEGATIVE_INFINITY 。
// 2）调用递归函数 dfs(root) 。
// 3）若 当前值（即 key ）出现的次数 val 与 resMaxCount，
// 则 将 val 放入 resList 中。
// 4）返回结果 resList 。
var findFrequentTreeSum = function(root) {
    const dfs = (curRoot = null) => {
        // 1）递归出口。
        if (!curRoot) {
            return 0;
        }

        // 2）递归主体。
        const {val, left, right} = curRoot,
            sum = val + dfs(left) + dfs(right);
        
        // 2.1）不断更新 resMap、resMaxCount 的值。
        resMap.set(sum, (resMap.get(sum) || 0) + 1);
        resMaxCount = Math.max(resMaxCount, resMap.get(sum));

        // 2.2）返回当前 sum 值！！
        return sum;
    };

    // 1）状态初始化：esMap = new Map(), resMaxCount = Number.NEGATIVE_INFINITY 。
    let resMap = new Map(),
        resMaxCount = Number.NEGATIVE_INFINITY;

    // 2）调用递归函数 dfs(root) 。
    dfs(root);

    // 3）若 当前值（即 key ）出现的次数 val 与 resMaxCount，
    // 则 将 val 放入 resList 中。
    let resList = [];
    for (const [key, val] of resMap) {
        if (val === resMaxCount) {
            resList.push(key);
        }
    }

    // 4）返回结果 resList 。
    return resList;
};

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