LEETCODE: Sum Root to Leaf Numbers

停止更新+1

于 2015-01-03 16:15:40 发布

阅读量319

点赞数

CC 4.0 BY-SA版权

文章标签： LEETCODE 算法递归

本文链接：https://blog.youkuaiyun.com/anyicheng2015/article/details/42364159

LEETCODE 专栏收录该内容

153 篇文章

订阅专栏

本文介绍了一种算法，用于求解给定二叉树中所有从根节点到叶子节点的路径所代表数字的总和。通过深度优先搜索遍历二叉树，实现了对每个路径数值的有效计算。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sum;
    void depthSearch(TreeNode *root, int num) {
        if(root->left == NULL && root->right == NULL)   sum += num;
        if(root->left) depthSearch(root->left, num * 10 + root->left->val);
        if(root->right) depthSearch(root->right, num * 10 + root->right->val);
    }
    int sumNumbers(TreeNode *root) {
        if(root == NULL)    return 0;
        depthSearch(root, root->val);
        return sum;
    }
};