从零开始的LC刷题(62): Binary Tree Paths 二叉树路径_c语言将二叉树根节点到p所指向节点路径上的所有节点依次压入栈中-优快云博客

本文详细解析了一道关于二叉树的算法题目，通过递归和迭代两种方式实现了寻找并输出所有从根节点到叶子节点的路径。文章提供了完整的C++代码实现，并对比了两种方法的性能和内存消耗。

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原题：

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

题意是求出所有二叉树到叶子节点的路径并且以string的形式输出，典型的先序遍历，最蠢的是你还要用'->'表示相连，正好看了stringstream转化int为stream可以利用一下，另外由于循环调用压栈时不好记忆当前路径（其实也行不过还需要一个vector容器，浪费不少空间），就使用递归了，结果：

Success

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Binary Tree Paths.

Memory Usage: 12 MB, less than 45.45% of C++ online submissions for Binary Tree Paths.

代码：

/**
 * 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> ans;
    vector<string> binaryTreePaths(TreeNode* root) {
        if(root==nullptr){return ans;}
        btp(root,"");
        return ans;
    }
    void btp(TreeNode* root,string path){
        stringstream ss;
        ss<<root->val;
        string temp;
        ss>>temp;
        path+=temp;
        if(root->left==nullptr&&root->right==nullptr){
            ans.push_back(path);
        }
        path+='-';
        path+='>';
        if(root->left!=nullptr){btp(root->left,path);}
        if(root->right!=nullptr){btp(root->right,path);}
    }
};

为了减少内存消耗，还是写了循环调用。循环需要三个vector，一个用来返回结果，一个用来压栈时记忆路径，一个用来压栈，结果：

Success

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Binary Tree Paths.

Memory Usage: 9.8 MB, less than 99.55% of C++ online submissions for Binary Tree Paths.

代码：

/**
 * 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) {
        vector<string> ans;
        if(root==nullptr){return ans;}
        vector<string> path;
        vector<TreeNode*> lib;
        path.push_back("");
        lib.push_back(root);
        while(!lib.empty()){
            TreeNode *cur=lib.back();
            lib.pop_back();
            string temp=path.back();
            path.pop_back();
            stringstream ss;
            ss<<cur->val;
            string s;
            ss>>s;
            temp+=s;
            if(cur->left==nullptr&&cur->right==nullptr){
                ans.push_back(temp);
                continue;
            }
            temp+="->";
            if(cur->left!=nullptr){
                lib.push_back(cur->left);
                path.push_back(temp);
            }
            if(cur->right!=nullptr){
                lib.push_back(cur->right);
                path.push_back(temp);
            }
            
        }
        return ans;
    }
};