Path Sum

二叉树路径求和问题

最新推荐文章于 2024-10-08 17:25:07 发布

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本文探讨了一个经典的二叉树问题：如何判断从根节点到叶子节点是否存在一条路径，使得路径上的节点值之和等于给定的数值。通过递归深度优先搜索(DFS)的方法，我们能够有效地解决这一问题。

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同时更新sum值，如果sum为0则表示成功返回。

/**
 * Definition for binary tree
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if (root == NULL) return false;
        sum -= root->val;
        if (root->left == NULL && root->right == NULL) {
            return sum == 0;
        }
        
        return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
    }
};