[leetcode 113] Path Sum II

最新推荐文章于 2025-12-28 17:52:30 发布

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本文介绍了一种算法，该算法接收一个二叉树及一个目标和作为输入，然后找出所有从根节点到叶子节点的路径，这些路径上的节点值之和等于给定的目标和。例如，在给定的二叉树中寻找和为22的路径。

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]
]

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