解题思路

path用来保存路径

result保存要返回的字符串

使用前序遍历二叉树，将路过的路径保存到path。

代码

Python

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
        result = []
        s = ""
        def dfs(root, s):
            if root is None:
                return
            s += str(root.val) + "->"
            if (not root.left) and (not root.right):
                result.append(s[:len(s) - 2])
                return
            dfs(root.left, s)
            dfs(root.right, s)
        dfs(root, s)
        return result

C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<int> path;
        vector<string> result;
        if (root == NULL) return result;
        traversal(root, path, result);
        return result;
    }
    void traversal(TreeNode* cur, vector<int> &path, vector<string> &result){
        path.push_back(cur->val);
        // 叶子节点
        if (cur->left == NULL && cur->right == NULL){
            string sPath;
            for (int i = 0; i < path.size() - 1; ++i){
                sPath += to_string(path[i]);
                sPath += "->";
            }
            sPath += to_string(path[path.size() - 1]);
            result.push_back(sPath);
            return;
        }
        // 下一个节点不为空时
        if (cur->left) {
            traversal(cur->left, path, result);
            path.pop_back();
        }
        if (cur->right) {
            traversal(cur->right, path, result);
            path.pop_back();
        }
    }
};

Go

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
var paths []string

func binaryTreePaths(root *TreeNode) []string {
    paths = []string{}
    constructPaths(root, "")
    return paths
}

func constructPaths(root *TreeNode, path string) {
    if root != nil {
        pathSB := path
        pathSB += strconv.Itoa(root.Val)
        if root.Left == nil && root.Right == nil {
            paths = append(paths, pathSB)
        } else {
            pathSB += "->"
            constructPaths(root.Left, pathSB)
            constructPaths(root.Right, pathSB)
        }
    }
}