LeetCode OJ -- Binary Tree Paths

最新推荐文章于 2025-08-12 12:49:47 发布

ubooksapp

最新推荐文章于 2025-08-12 12:49:47 发布

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文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/ubooksapp/article/details/47836635

本文介绍了一种用于查找二叉树中所有从根节点到叶子节点路径的递归算法。通过递归地构建路径并将当前节点添加到路径中，最终得到所有可能的路径集合。

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标签（空格分隔）： LeetCode OJ BinaryTree

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

For example, given the following binary tree:
   1
 /   \
2     3
 \
  5
All root-to-leaf paths are:

[“1->2->5”, “1->3”]

No doubt that we should use recursive method。

/**
 * 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<string> binaryTreePaths(TreeNode* root) {
        if (root == NULL){
            vector<string> empty;
            return empty;
        } 

        if (root->left == NULL && root->right == NULL){ // leaf
            vector<string> path;
            path.push_back(std::to_string(root->val));
            return path;
        }

        vector<string> allPaths = combine(binaryTreePaths(root->left), binaryTreePaths(root->right)); // combine all subpaths

        vector<string> result;
        for (vector<string>::iterator it = allPaths.begin(); it != allPaths.end(); ++it){
            result.push_back(std::to_string(root->val) + "->" + *it); // add current node to the paths
        }

        return result;
    }

    vector<string> combine(vector<string> left, vector<string> right){ // this is how to combine two vectors
        vector<string> result;
        result.reserve(left.size() + right.size() ); // preallocate memory
        result.insert( result.end(), left.begin(), left.end() );
        result.insert( result.end(), right.begin(), right.end() );
        return result;
    }
};