[LeetCode] Path Sum

最新推荐文章于 2025-09-10 14:50:50 发布

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最新推荐文章于 2025-09-10 14:50:50 发布

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CC 4.0 BY-SA版权

文章标签： c/c++

原文链接：http://www.cnblogs.com/chkkch/archive/2012/11/13/2767746.html

本文探讨了如何确定一棵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目标值。通过深度优先搜索(DFS)实现了这一功能，并提供了详细的代码实现。

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.

DFS或者BFS。同时，当本节点是叶子节点才判断和，否则节点为空时返回false

 1 /**
 2  * Definition for binary tree
 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 dfs(TreeNode *node, int sum, int curSum)
13     {
14         if (node == NULL)
15             return false;
16         
17         if (node->left == NULL && node->right == NULL)
18             return curSum + node->val == sum;
19                
20         return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val);
21     }
22     
23     bool hasPathSum(TreeNode *root, int sum) {
24         // Start typing your C/C++ solution below
25         // DO NOT write int main() function
26         return dfs(root, sum, 0);
27     }
28 };