【二叉树】257. Binary Tree Paths

最新推荐文章于 2021-12-30 10:05:52 发布

原创最新推荐文章于 2021-12-30 10:05:52 发布 · 160 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种算法，用于找出二叉树中所有从根节点到叶子节点的路径。通过深度优先搜索(DFS)和广度优先搜索(BFS)两种方法实现，并附带详细的代码示例。

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

For example, given the following binary tree:

All root-to-leaf paths are:

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

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

一道和深搜、树有关的题

注意to_string的用法，比较棘手的地方是字符串必须最后压进ans里：

/**
 * 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)return ans;
        dfs(ans,root,to_string(root->val));
        return ans;
    }
    void dfs(vector<string>& ans,TreeNode* root,string str){
        if(!root->left&&!root->right){
            ans.push_back(str);
        }
        if(root->left)dfs(ans,root->left,str+"->"+to_string(root->left->val));
        if(root->right)dfs(ans,root->right,str+"->"+to_string(root->right->val));
    }
};

用广度优先也可以，需要两个队列：

/**
 * 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) {
        queue<TreeNode*> q;
        queue<string> str_q;
        vector<string> ans;
        
        if(!root)return ans;
        
        q.push(root);//enqueue
        str_q.push("");//enqueue
        
        while(!q.empty()){
            TreeNode* helper=q.front();
            string tmp=str_q.front();
            q.pop();//dequeue
            str_q.pop();//dequeue
            
            if((!helper->left)&&(!helper->right))
                ans.push_back(tmp+to_string(helper->val));
            if(helper->left){
                q.push(helper->left);
                str_q.push(tmp+to_string(helper->val)+"->");
            }
            if(helper->right){
                q.push(helper->right);
                str_q.push(tmp+to_string(helper->val)+"->");
            }
        }
        return ans;
    }
};

广度优先的话这样会更清晰：

/**
 * 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) {
        queue<TreeNode*> q;
        queue<string> str_q;
        vector<string> ans;
        
        if(!root)return ans;
        
        q.push(root);//enqueue
        str_q.push(to_string(root->val));//enqueue
        
        while(!q.empty()){
            TreeNode* helper=q.front();
            string tmp=str_q.front();
            q.pop();//dequeue
            str_q.pop();//dequeue
            
            if((!helper->left)&&(!helper->right))
                ans.push_back(tmp);
            if(helper->left){
                q.push(helper->left);
                str_q.push(tmp+"->"+to_string(helper->left->val));
            }
            if(helper->right){
                q.push(helper->right);
                str_q.push(tmp+"->"+to_string(helper->right->val));
            }
        }
        return ans;
    }
};