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LeetCode #1302 Deepest Leaves Su

LeetCode #1302 Deepest Leaves Su

作者: air_melt | 来源:发表于2022-08-28 23:34 被阅读0次

    1302 Deepest Leaves Sum 层数最深叶子节点的和

    Description:

    Given the root of a binary tree, return the sum of values of its deepest leaves.

    Example:

    Example 1:

    [图片上传失败...(image-3f9437-1661700880171)]

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

    Example 2:

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

    Constraints:

    The number of nodes in the tree is in the range [1, 10^4].
    1 <= Node.val <= 100

    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:

    示例 1:

    [图片上传失败...(image-8e9b95-1661700880171)]

    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, 10^4] 之间。
    1 <= Node.val <= 100

    思路:

    1. 迭代
      使用队列层序遍历
      每到新的一层, 重置累加值为 0
      加上该层的所有节点的值, 直到最后一层
      时间复杂度为 O(n), 空间复杂度为 O(n)
    2. 递归
      记录最大层数和结果
      每次更新最大层时, 将该节点的值赋给结果
      累计同一层的所有节点的值
      时间复杂度为 O(n), 空间复杂度为 O(n)

    代码:

    C++:

    /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
     *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
     * };
     */
    class Solution 
    {
    public:
        int deepestLeavesSum(TreeNode* root) 
        {
            int result = 0;
            queue<TreeNode*> q;
            if (root) q.push(root);
            while (!q.empty())
            {
                result = 0;
                for (int i = 0, s = q.size(); i < s; i++)
                {
                    TreeNode* cur = q.front();
                    q.pop();
                    result += cur -> val;
                    if (cur -> left) q.push(cur -> left);
                    if (cur -> right) q.push(cur -> right);
                }
            }
            return result;
        }
    };
    

    Java:

    /**
     * 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 {
        private int result = 0, maxLevel = 0;
        
        public int deepestLeavesSum(TreeNode root) {
            helper(root, 0);
            return result;
        }
        
        private void helper(TreeNode root, int level) {
            if (root == null) return;
            if (level > maxLevel) {
                maxLevel = level;
                result = root.val;
            } else if (level == maxLevel) result += root.val;
            helper(root.left, level + 1);
            helper(root.right, level + 1);
        }
    }
    

    Python:

    # Definition for a binary tree node.
    # class TreeNode:
    #     def __init__(self, val=0, left=None, right=None):
    #         self.val = val
    #         self.left = left
    #         self.right = right
    class Solution:
        def deepestLeavesSum(self, root: Optional[TreeNode]) -> int:
            result, queue = 0, [root] if root else []
            while queue:
                result, s = 0, len(queue)
                for _ in range(s):
                    cur = queue.pop(0)
                    result += cur.val
                    if cur.left:
                        queue.append(cur.left)
                    if cur.right:
                        queue.append(cur.right)
            return result
    

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