层数最深叶子节点的和

1302. 层数最深叶子节点的和

难度中等47收藏分享切换为英文接收动态反馈

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。

广度优先遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int deepestLeavesSum(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            boolean flag = false;
            int sum = 0;
            for (int i = 0; i < size; i++) {
                root = queue.poll();
                sum += root.val;
                if (root.left != null) {
                    flag = true;
                    queue.offer(root.left);
                }
                if (root.right != null) {
                    flag = true;
                    queue.offer(root.right);
                }
            }
            if (!flag) return sum;
        }
        return 0;
    }
}

深度优先遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private int maxLevel = 0;
    private int sum = 0;
    public int deepestLeavesSum(TreeNode root) {
        if(root == null) return 0;
        helper(root, 0);
        return sum;
    }

    public void helper(TreeNode root, int curLevel) {
        if(root == null) return;
        if(root.left == null && root.right == null) {
            if(curLevel == maxLevel) {
                sum += root.val;
            }else if(curLevel > maxLevel) {
                maxLevel = curLevel;
                sum = root.val;
            }
        }
        helper(root.left, curLevel + 1);
        helper(root.right, curLevel + 1);
    }

}

赏

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