LeetCode——1302. 层数最深叶子节点的和

题目描述：

给你一棵二叉树的根节点 root ，请你返回层数最深的叶子节点的和 。

示例 1：

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

输入：root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
输出：19

提示：

• 树中节点数目在范围 [1, 104] 之间。
• 1 <= Node.val <= 100

dfs代码如下：

/**
 * 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 {
    public int sum = 0;

    public int deepestLeavesSum(TreeNode root) {
        int maxDepth = depth(root);
        dfs(root, 1, maxDepth);
        return sum;
    }

    public void dfs(TreeNode root, int depth, int maxDepth) {
        if (root == null) {
            return;
        }
        if (depth == maxDepth) {
            sum += root.val;
        }
        dfs(root.left, depth + 1, maxDepth);
        dfs(root.right, depth + 1, maxDepth);
    }

    public int depth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return Math.max(depth(root.left), depth(root.right)) + 1;
    }
}

bfs代码如下：

/**
 * 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 {
    public int deepestLeavesSum(TreeNode root) {
        if (root == null) {
            return 0;
        }
        Queue<TreeNode> que = new LinkedList<>();
        que.offer(root);
        int ans = 0;
        while (!que.isEmpty()) {
            List<Integer> r = new ArrayList<>();
            int size = que.size();
            ans = 0;
            for (int i = 0; i < size; i++) {
                TreeNode t = que.poll();
                ans += t.val;
                if (t.left != null) {
                    que.offer(t.left);
                }
                if (t.right != null) {
                    que.offer(t.right);
                }
            }
        }
        return ans;
    }
}

执行结果：