LintCode 376. Binary Tree Path Sum

最新推荐文章于 2024-01-17 15:40:17 发布

纸上得来终觉浅绝知此事要躬行

最新推荐文章于 2024-01-17 15:40:17 发布

阅读量252

点赞数

CC 4.0 BY-SA版权

分类专栏： algorithm-design 文章标签： LintCode

本文链接：https://blog.youkuaiyun.com/roufoo/article/details/89762092

algorithm-design 专栏收录该内容

817 篇文章

订阅专栏

博客围绕二叉树路径和问题展开，给定二叉树，需找出节点和等于给定目标值的所有路径，路径需从根节点到叶节点。文中给出示例展示不同输入下的输出情况，还介绍了使用DFS（深度优先搜索）的解法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Binary Tree Path Sum
中文English
Given a binary tree, find all paths that sum of the nodes in the path equals to a given number target.

A valid path is from root node to any of the leaf nodes.

Example
Example 1:

Input:
{1,2,4,2,3}
5
Output: [[1, 2, 2],[1, 4]]
Explanation:
The tree is look like this:
1
/
2 4
/
2 3
For sum = 5 , it is obviously 1 + 2 + 2 = 1 + 4 = 5
Example 2:

Input:
{1,2,4,2,3}
3
Output: []
Explanation:
The tree is look like this:
1
/
2 4
/
2 3
Notice we need to find all paths from root node to leaf nodes.
1 + 2 + 2 = 5, 1 + 2 + 3 = 6, 1 + 4 = 5
There is no one satisfying it.

解法1：DFS
代码如下：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */


class Solution {
public:
    /*
     * @param root: the root of binary tree
     * @param target: An integer
     * @return: all valid paths
     */
    vector<vector<int>> binaryTreePathSum(TreeNode * root, int target) {
        if (!root) return vv;
        vector<int> v;
        helper(root, target, v);
        return vv;
    }

private:
    vector<vector<int>> vv;
    void helper(TreeNode * root, int target, vector<int> & v) {
        if (!root) return;
        if (target == root->val && !root->left && !root->right) {
            v.push_back(target);
            vv.push_back(v);
            v.pop_back();
            return;
        }
        v.push_back(root->val);
        helper(root->left, target - root->val, v);
        helper(root->right, target - root->val, v);
        v.pop_back();
    }
};