LeetCode 257. Binary Tree Paths

原题链接在这里：https://leetcode.com/problems/binary-tree-paths/

题目：

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"]

题解：

DFS 依次添加，终止条件是叶子节点.

Note:1. If use StringBuilder, remember backtracking before each return.

e.g. [1,2,3], 会写成["1->2","1->23"]. 因为扫过点2，StringBuilder 会变成1->2, 扫过点3，原来的点2没有去掉，就会变成1->23,但是第一个结果没有变成1->23是因为，用到sb.toString()时自动做了copy by value.

2 . 但如果使用String，在递归调用时可以直接使用str, 因为String 和 int, float一样是priority type, 是copy by value，而不像array等object 是 copy by reference.

Time Complexity: O(n), 这是dfs.

Space: O(h). h是树的高度，一共用了h层stack. StringBuilder size.

AC  Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        
        dfs(root, new StringBuilder(), res);
        return res;
    }
    
    private void dfs(TreeNode root, StringBuilder sb, List<String> res){
        if(root == null){
            return;
        }
        
        int len = sb.length();
        if(root.left == null && root.right == null){
            sb.append(root.val);
            res.add(sb.toString());
            sb.setLength(len);
            return;
        }
        
        sb.append(root.val).append("->");
        dfs(root.left, sb, res);
        dfs(root.right, sb, res);
        sb.setLength(len);
    }
}

上面使用StringBuilder, 下面使用String.

/**
 * 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> res = new ArrayList<String>();
        if(root == null){
            return res;
        }
        dfs(root, "", res);
        return res;
    }
    private void dfs(TreeNode root, String str, List<String> res){
        if(root.left == null && root.right == null){
            str = str + root.val;
            res.add(str);
        }
        str = str + root.val + "->";
        if(root.left != null){
            dfs(root.left, str, res);
        }
        if(root.right != null){
            dfs(root.right, str, res);
        }
    }
}

AC C++:

/**
 * 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<string> res;
        if(!root){
            return res;
        }

        dfs(root, "", res);
        return res;
    }

private:
    void dfs(TreeNode* root, string item, vector<string>& res){
        if(!root){
            return;
        }

        if(!root->left && !root->right){
            item += to_string(root->val);
            res.push_back(item);
            return;
        }

        item += to_string(root->val) + "->";
        dfs(root->left, item, res);
        dfs(root->right, item, res);
    }
};

类似Path Sum, Smallest String Starting From Leaf, Sum Root to Leaf Numbers.