LeetCode 0112 - Path Sum

Reck Zhang

发布于 2021-08-11 10:55:04

2030

发布于 2021-08-11 10:55:04

文章被收录于专栏：Reck Zhang

Path Sum

Desicription

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.

Solution

/**
 * 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 {
private:
    bool flag = 0;
    void searchTree(TreeNode* root, int sum, const int& target) {
        if(root == NULL || flag)
            return ;
        if(root->left == NULL && root->right == NULL)
            if(sum + root->val == target)
                flag = 1;
        searchTree(root->left, sum + root->val, target);
        searchTree(root->right, sum + root->val, target);
    }
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if(root == NULL)
            return 0;
        searchTree(root, 0, sum);
        return flag;
    }
};

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原始发表：2017-12-17，如有侵权请联系 cloudcommunity@tencent.com 删除

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