【算法千题案例】⚡️每日LeetCode打卡⚡️——64. 二叉树的所有路径

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
public IList<string> BinaryTreePaths(TreeNode root)
    {
        List<string> res = new List<string>();
        Backtrack257(root, root.val.ToString(), res);
        return res;
    }
    private void Backtrack257(TreeNode root, string path, List<string> res)
    {
        if (root.left == null && root.right == null)
        {
            res.Add(path);
            return;
        }
        if (root.left != null)
            Backtrack257(root.left, path + "->" + root.left.val.ToString(), res);
        if (root.right != null)
            Backtrack257(root.right, path + "->" + root.right.val.ToString(), res);
    }
}

/**
 * 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> paths = new ArrayList<String>();
        constructPaths(root, "", paths);
        return paths;
    }
    public void constructPaths(TreeNode root, String path, List<String> paths) {
        if (root != null) {
            StringBuffer pathSB = new StringBuffer(path);
            pathSB.append(Integer.toString(root.val));
            if (root.left == null && root.right == null) {  // 当前节点是叶子节点
                paths.add(pathSB.toString());  // 把路径加入到答案中
            } else {
                pathSB.append("->");  // 当前节点不是叶子节点，继续递归遍历
                constructPaths(root.left, pathSB.toString(), paths);
                constructPaths(root.right, pathSB.toString(), paths);
            }
        }
    }
}