原题如下
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,
代码语言：javascript复制 5
/
4 8
/ /
11 13 4
/ /
7 2 5 1Return:
代码语言：javascript复制[
[5,4,11,2],
[5,8,4,5]
]思路
是上一篇博客 LeetCode112. Path Sum 的升级版，用同样的递归思想，只不过这次要从左右子树接收路径数组，并在每一条可行路径前插入根节点的值，以形成一条最终完整的路径。
代码
代码语言：javascript复制/**
* 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>> paths;
if(root == NULL) return paths;
if(root->left == NULL && root->right == NULL){
if(root->val == sum){
vector<int> path;
path.push_back(root->val);
paths.push_back(path);
return paths;
}
}
vector<vector<int>> leftPaths = pathSum(root->left, sum - root->val);
vector<vector<int>> rightPaths = pathSum(root->right, sum - root->val);
// 向所有路径的头部加入根节点的值
for(int i = 0; i < leftPaths.size(); i )
leftPaths[i].insert(leftPaths[i].cbegin(), root->val);
for(int i = 0; i < rightPaths.size(); i )
rightPaths[i].insert(rightPaths[i].cbegin(), root->val);
// 合并所有路径
leftPaths.insert(leftPaths.cend(), rightPaths.cbegin(), rightPaths.cend());
return leftPaths;
}
};
