113. Path Sum II

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

Note: A leaf is a node with no children.

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]
]

思路

前一题和这一题是dfs的两种不同问题，Path Sum是求是否存在解，这一题是求所有解，所以很简单，把所有的解都记录下来即可。

C++11

/**
 * Definition for a binary tree node.
 * 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;
        dfs(root, sum, vector<int>{}, res);
        return res;
    }

    void dfs(TreeNode* root, int sum, vector<int> path, vector<vector<int>> &res) {
        if (root == NULL) return;
        path.emplace_back(root->val);
        if (root->left == NULL and root->right == NULL and root->val == sum) res.emplace_back(path);
        if (root->left) dfs(root->left, sum - root->val, path, res);
        if (root->right) dfs(root->right, sum - root->val, path, res);
    }
};