LeetCode OJ：Path Sum（路径之和）-优快云博客

本文介绍了一种通过递归方法判断二叉树中是否存在从根节点到叶节点的路径，使得路径上所有节点值之和等于给定值的方法。

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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.

For example:
Given the below binary tree and sum = 22,

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

要求求出是否有一条路径和与给出的值相等，注意中间节点与叶子节点的判断：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool hasPathSum(TreeNode* root, int sum) {
13         if(root == NULL) return false;
14             return checkSum(root, sum);
15     }
16     
17     bool checkSum(TreeNode* root, int sum)
18     {
19         if(root != NULL && sum == root->val && root->left == NULL && root->right == NULL){
20             return true;//上面这个判断确实是叶子节点，值也同时满足
21         }
22         else if(root == NULL)
23             return false;
24         else
25             return checkSum(root->left, sum - root->val) || checkSum(root->right, sum - root->val);    
26     }
27 };

java版本的如下，递归版本的没上面那么麻烦：

 1 public class Solution {
 2     public boolean hasPathSum(TreeNode root, int sum) {
 3         if(root == null)
 4             return false;
 5         if(root.left == null && root.right == null){
 6             return sum == root.val;
 7         }
 8         return hasPathSum(root.left, sum - root.val) || 
 9                hasPathSum(root.right, sum - root.val);
10     }
11 }