leetcode257. All paths of the binary tree/bfs,dfs

Subject: 257. All paths of a binary tree

Given a binary tree, return all paths from the root node to the leaf nodes.

Explanation: A leaf node refers to a node without child nodes.

Example:

输入:

   1
 /   \
2     3
 \
  5

输出: ["1->2->5", "1->3"]

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

Source: LeetCode
Link: https://leetcode-cn.com/problems/binary-tree-paths
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Basic idea: dfs

/**
 * 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> res;
    vector<string> binaryTreePaths(TreeNode* root) {
    
    
        if(root == NULL)
            return res;        
        dfs(root, "");
        return res;
    }
    void dfs(TreeNode * root, string cur){
    
    
        string temp = to_string(root->val);
        if(cur == ""){
    
                
            cur += temp;
        }
        else{
    
    
            cur += "->" + temp;
        }

        if(root->left == NULL && root->right == NULL){
    
    //遇到根节点存放结果
            res.push_back(cur);
            return;
        }
        if(root->left){
    
    
            dfs(root->left, cur);
        }
        if(root->right){
    
    
            dfs(root->right, cur);
        }
        

    }
};

Basic idea 2: bfs

You can also use bfs

  • Maintain two queues, one queue is used to store the current node, and one queue is used to store the path formed by the node.
  • Every time a node is ejected from the queue, a path is also ejected at the same time,
  • If the current node has a left child, the left child is connected to the current path and the path is enqueued; similarly, if the current node has children, the right child of the current node is also added to the current path and the path is enqueued.
/**
 * 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> res;
        if(root == NULL)
            return res;
        queue<TreeNode*> q;
        queue<string> path;
        q.push(root);
        path.push(to_string(root->val));
        while(!q.empty()){
    
    
            auto temp_n = q.front();
            auto temp_p = path.front();
            q.pop();
            path.pop();
            if(temp_n->left == NULL && temp_n->right == NULL){
    
    
                res.push_back(temp_p);
            }
            if(temp_n->left){
    
    
                q.push(temp_n->left);
                path.push(temp_p + "->" + to_string(temp_n->left->val));
            }
            if(temp_n->right){
    
    
                q.push(temp_n->right);
                path.push(temp_p + "->" + to_string(temp_n->right->val));
            }
        }
        return res;
    }
};

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Origin blog.csdn.net/qq_31672701/article/details/108407721
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