【Leetcode】Path Sum II

最新推荐文章于 2017-06-29 10:05:00 发布

原创最新推荐文章于 2017-06-29 10:05:00 发布 · 1k 阅读

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本文介绍了一种使用深度优先搜索(DFS)算法解决二叉树中所有根到叶子节点路径之和等于指定值的问题的方法。通过递归地遍历二叉树，当到达叶子节点且路径和等于目标值时，将该路径保存下来。

题目：

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]
]

解题思路：用DFS算法进行搜索，搜索到叶子节点并且路径和等于设定值时，将路径存入容器。

代码：

/**
 * 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>> Path;
        vector<int> CurrPath;
        build_path(Path, CurrPath, root, sum);
        return Path;
    }
private:
    void build_path(vector<vector<int>> &Path,vector<int> &CurrPath, TreeNode *root, int sum){
        if(!root)return;
        CurrPath.push_back(root->val);
        if(!root->left&&!root->right){
            if(root->val==sum)Path.push_back(CurrPath);
        }else{
            build_path(Path,CurrPath,root->left,sum-root->val);
            build_path(Path,CurrPath,root->right,sum-root->val);
        }
        CurrPath.pop_back();
    }
};