Path Sum

最新推荐文章于 2025-08-21 20:43:22 发布

原创最新推荐文章于 2025-08-21 20:43:22 发布 · 256 阅读

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一、问题描述

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.

二、思路

这题有点小坑，开始我写的代码的这一句

        if(root -> val == sum && root -> left ==NULL &&root -> right ==NULL) return true;

没有加

&& root -> left ==NULL &&root -> right ==NULL

导致无法通过，原因是：只有当左右子树为空的情况下，即只有一个根节点的情况且根节点数据域的值和sum值相等，才返回true，只要有子树非空即使根节点数据域的值和sum值相等，也返回false。

三、代码

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