[LeetCode] 257. Binary Tree Paths 二叉树路径

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"]

给一个二叉树，返回所有根到叶节点的路径。

Java:

/** 
 * Definition for a binary tree node. 
 * public class TreeNode { 
 *     int val; 
 *     TreeNode left; 
 *     TreeNode right; 
 *     TreeNode(int x) { val = x; } 
 * } 
 */  
public class Solution {  
   public List<String> binaryTreePaths(TreeNode root) {  
        List<String> list = new ArrayList<>();  
        binaryTreePathsHelper(root, list, new String());  
        return list;  
    }  
  
    public void binaryTreePathsHelper(TreeNode root, List<String> list, String string) {  
        if (root == null) {  
            return;  
        }  
        if (root.left == null && root.right == null) {  
                string = string + root.val;  
            list.add(string);  
            return;  
        }  
  
        binaryTreePathsHelper(root.left, list, string + root.val + "->");  
        binaryTreePathsHelper(root.right, list, string + root.val + "->");  
    }  
}　

Python:

class Solution:
    # @param {TreeNode} root
    # @return {string[]}
    def binaryTreePaths(self, root):
        result, path = [], []
        self.binaryTreePathsRecu(root, path, result)
        return result
    
    def binaryTreePathsRecu(self, node, path, result):
        if node is None:
            return

        if node.left is node.right is None:
            ans = ""
            for n in path:
                ans += str(n.val) + "->"
            result.append(ans + str(node.val))

        if node.left:
            path.append(node)
            self.binaryTreePathsRecu(node.left, path, result)
            path.pop()

        if node.right:
            path.append(node)
            self.binaryTreePathsRecu(node.right, path, result)
            path.pop()

C++: DFS

class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        if (root) dfs(root, "", res);
        return res;
    }
    void dfs(TreeNode *root, string out, vector<string> &res) {
        out += to_string(root->val);
        if (!root->left && !root->right) res.push_back(out);
        else {
            if (root->left) dfs(root->left, out + "->", res);
            if (root->right) dfs(root->right, out + "->", res);
        }
    }
};

C++:

class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        if (!root) return {};
        if (!root->left && !root->right) return {to_string(root->val)};
        vector<string> left = binaryTreePaths(root->left);
        vector<string> right = binaryTreePaths(root->right);
        left.insert(left.end(), right.begin(), right.end());
        for (auto &a : left) {
            a = to_string(root->val) + "->" + a;
        }
        return left;
    }
};

　

类似题目：

[LeetCode] 112. Path Sum 路径和

[LeetCode] 113. Path Sum II 路径和 II

[LeetCode] 437. Path Sum III 路径和 III  

　

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