257. Binary Tree Paths_binarytreepaths 257-优快云博客

本文链接：https://blog.youkuaiyun.com/brynjiang/article/details/102632994

本文探讨了如何使用遍历法和分治法解决二叉树的所有根到叶路径问题，通过具体代码示例展示了两种方法的实现细节，强调了避免在root为空时将结果误加入列表的重要性。

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题目

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

我的想法

遍历法，需要注意不能在root == null时把结果加入list，这样会重复计算，而且并没有走到真正的叶子结点(缺少一边的子树也会输出，但实际上这并不是叶子结点)

public class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> list = new LinkedList<>();
        if(root == null) {
            return list;
        }
        helper("", root, list);
        return list;
    }
    
    private void helper(String path, TreeNode root, List<String> list) {
        if(root == null) {
            return;
        }
        if(root.right == null && root.left == null) {
            path += root.val;
            list.add(path);
            return;
        }
        
        path += root.val;
        helper(path + "->", root.left, list);
        helper(path + "->", root.right, list);
    }
}

解答

jiuzhang solution: divide
一开始并没有想到怎样用分治来做，主要是智障的觉得从后往前路径会是反的，path是String直接在前方插入就好了啊！！！！
对于分治法来说可以直接做到root == null，这个有点像dp的bottom up，由若干个小数组(即路径)向上汇总成一个大数组

public class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> paths = new LinkedList<>();
        if(root == null) {
            return paths;
        }
        //store leftpaths, rightpaths
        List<String> leftpaths = binaryTreePaths(root.left);
        List<String> rightpaths = binaryTreePaths(root.right);
        
        //add cur node to the leftpaths, rightpaths 
        //and integrate them to the cur node paths
        for(String path : leftpaths) {
            paths.add(root.val + "->" + path);
        }
        for(String path : rightpaths) {
            paths.add(root.val + "->" + path);
        }
        
        if(paths.size() == 0) {
            paths.add("" + root.val);
        }
        return paths;
    }
}

jiuzhang solution: traversal
想法一样，写的更漂亮一些

public class Solution {
    /**
     * @param root the root of the binary tree
     * @return all root-to-leaf paths
     */
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new ArrayList<String>();
        if (root == null) {
            return result;
        }
        helper(root, String.valueOf(root.val), result);
        return result;
    }
    
    private void helper(TreeNode root, String path, List<String> result) {
        if (root == null) {
            return;
        }
        
        if (root.left == null && root.right == null) {
            result.add(path);
            return;
        }
        
        if (root.left != null) {
            helper(root.left, path + "->" + String.valueOf(root.left.val), result);
        }
        
        if (root.right != null) {
            helper(root.right, path + "->" + String.valueOf(root.right.val), result);
        }
    }
}