LEETCODE: Path Sum II

最新推荐文章于 2025-11-25 00:06:26 发布

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#LEETCODE #算法 #递归 #二叉树 #遍历

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本文介绍了一种算法，用于在给定的二叉树中找到所有从根节点到叶子节点的路径，使得这些路径上的元素之和等于指定的数值。通过实例演示了如何使用该算法，并提出了在完成此轮LeetCode练习后考虑非递归方法的建议。

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and

sum
 = 22

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

return

[
   [5,4,11,2],
   [5,8,4,5]
]

等这一轮LEETCODE 刷完，下一轮考虑非递归的方法。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void pathSum(TreeNode *root, int sum, vector<vector<int> > &results, vector<int> current) {
        if(root == NULL) return;
        if(root->val == sum && root->left == NULL && root->right == NULL) {
            current.push_back(root->val);
            results.push_back(current);
            return;
        }
        
        current.push_back(root->val);
        pathSum(root->left, sum - root->val, results, current);
        pathSum(root->right, sum - root->val, results, current);
    }
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        vector<vector<int> > results;
        vector<int> current;
        pathSum(root, sum, results, current);
        return results;
    }
};