112. Path Sum(easy)-优快云博客

本文探讨了二叉树中是否存在从根节点到叶子节点的路径，使得路径上所有节点值之和等于给定的数值。通过递归深度优先搜索(DFS)算法，检查每个可能的路径，直至找到满足条件的路径或遍历完整棵树。

Easy

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Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

C++:

/*
 * @Author: SourDumplings
 * @Link: https://github.com/SourDumplings/
 * @Email: changzheng300@foxmail.com
 * @Description: https://leetcode.com/problems/path-sum/
 * @Date: 2019-03-08 21:38:19
 */

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution
{
  public:
    bool hasPathSum(TreeNode *root, int sum)
    {
        if (!root)
        {
            return false;
        }
        int nowSum = root->val;
        return dfs(root, nowSum, sum);
    }

    bool dfs(TreeNode *root, int nowSum, int sum)
    {
        if (!root->left && !root->right && nowSum == sum)
        {
            return true;
        }
        else if (root->left && dfs(root->left, nowSum + root->left->val, sum) ||
                 root->right && dfs(root->right, nowSum + root->right->val, sum))
        {
            return true;
        }
        return false;
    }
};