Leetcode 112. Path Sum (Easy) (cpp)

原创于 2016-07-23 18:39:36 发布 · 568 阅读

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本文介绍了解决LeetCode上题目112. Path Sum的方法。该题要求判断二叉树中是否存在一条从根节点到叶子节点的路径，使得路径上的节点值之和等于给定的目标值。文章提供了使用深度优先搜索(DFS)策略的C++代码实现。

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Leetcode 112. Path Sum (Easy) (cpp)

Tag: Tree, Depth-first Search

Difficulty: Easy

/*

112. Path Sum (Easy)

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.

For 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.

*/
/**
 * 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 == NULL) {
            return false;
        } else if (root->left == NULL && root->right == NULL && sum == root->val) {
            return true;
        }
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
};