LeetCode #257 - Binary Tree Paths_given a binary tree, return all root-to-leaf paths-优快云博客

本文链接：https://blog.youkuaiyun.com/LawFile/article/details/81189736

本文介绍了一种用于寻找二叉树所有从根节点到叶子节点路径的算法实现。通过递归方式遍历二叉树的每个节点，并记录下经过的路径。当到达叶子节点时，将该路径保存为最终结果之一。

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题目描述：

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

给定一个二叉树，返回从根节点到叶节点的所有路径。显然是利用递归，从根节点开始遍历，每次遍历的节点都加入遍历路径中，当遍历到叶节点则递归结束。

class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<int> path;
        vector<vector<int>> v;
        binaryTreePaths_recur(root,path,v);
        vector<string> result;
        for(int i=0;i<v.size();i++)
        {
            if(v[i].size()==1) result.push_back(to_string(v[i][0]));
            else
            {
                string s;
                for(int j=0;j<v[i].size();j++)
                {
                    if(j==v[i].size()-1) s+=to_string(v[i][j]);
                    else s+=to_string(v[i][j])+"->";
                }
                result.push_back(s);
            }
        }
        return result;
    }
    
    void binaryTreePaths_recur(TreeNode* root, vector<int> path, vector<vector<int>>& result)
    {
        if(root==NULL) return;
        path.push_back(root->val);
        if(root->left==NULL&&root->right==NULL)
        {
            result.push_back(path);
            return;
        }
        else
        {
            binaryTreePaths_recur(root->left,path,result);
            binaryTreePaths_recur(root->right,path,result);
        }
    }
};