1302-层数最深叶子节点的和

1302-层数最深叶子节点的和

给你一棵二叉树，请你返回层数最深的叶子节点的和。

示例：

输入：root = [1,2,3,4,5,null,6,7,null,null,null,null,8]
输出：15

提示：

• 树中节点数目在 1 到 10^4 之间。
• 每个节点的值在 1 到 100 之间。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/deepest-leaves-sum
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

DFS

    private int deepest = 0;

    private Map<Integer, Integer> map = new HashMap<>();

    public int deepestLeavesSum(TreeNode root) {
        scan(root, 0);
        return map.getOrDefault(deepest, 0);
    }

    private void scan(TreeNode root, int depth) {

        deepest = Math.max(deepest, depth);

        if(root.left == null && root.right == null) {
            map.put(depth, map.getOrDefault(depth, 0) + root.val);
        }

        if(root.left != null) {
            scan(root.left, depth + 1);
        }

        if(root.right != null) {
            scan(root.right, depth + 1);
        }
    }

改进：缩小map

    private int deepest = 0;

    private Map<Integer, Integer> map = new HashMap<>();

    public int deepestLeavesSum(TreeNode root) {
        scan(root, 0);
        return map.getOrDefault(deepest, 0);
    }

    private void scan(TreeNode root, int depth) {
        if(depth > deepest) {
            deepest = depth;
            // 说明之前存储的数据无用
            map.clear();
        }

        if(depth == deepest && root.left == null && root.right == null) {
            map.put(depth, map.getOrDefault(depth, 0) + root.val);
        }

        if(root.left != null) {
            scan(root.left, depth + 1);
        }

        if(root.right != null) {
            scan(root.right, depth + 1);
        }
    }

进一步改进：不使用map，直接用int全局变量存储结果

    private int deepest = 0;
    
    private int sum = 0;
    
    public int deepestLeavesSum(TreeNode root) {
        scan(root, 0);
        return sum;
    }

    private void scan(TreeNode root, int depth) {
        if(depth > deepest) {
            deepest = depth;
            sum = 0;
        }

        if(depth == deepest && root.left == null && root.right == null) {
            sum += root.val;
        }

        if(root.left != null) {
            scan(root.left, depth + 1);
        }

        if(root.right != null) {
            scan(root.right, depth + 1);
        }
    }

BFS

    public int deepestLeavesSum(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int res = 0;
        while (!queue.isEmpty()) {
            res = 0;
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode tmp = queue.poll();
                res += tmp.val;
                if (tmp.left != null)
                    queue.offer(tmp.left);
                if (tmp.right != null)
                    queue.offer(tmp.right);
            }
        }
        return res;
    }