[LeetCode] #257 二叉树的所有路径

给你一个二叉树的根节点 root ，按 任意顺序 ，返回所有从根节点到叶子节点的路径。

叶子节点 是指没有子节点的节点。

输入：root = [1,2,3,null,5]

输出：["1->2->5","1->3"]

递归，将list作为参数传递

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

递归，将list作为全局变量

class Solution {
    List<String> list = new ArrayList<>();

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

递归，将list作为返回值

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

迭代

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> paths = new ArrayList<String>();
        if (root == null) return paths;

        Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();
        Queue<String> pathQueue = new LinkedList<String>();

        nodeQueue.offer(root);
        pathQueue.offer(Integer.toString(root.val));

        while (!nodeQueue.isEmpty()) {
            TreeNode node = nodeQueue.poll(); 
            String path = pathQueue.poll();

            if (node.left == null && node.right == null) {
                paths.add(path);
            } else {
                if (node.left != null) {
                    nodeQueue.offer(node.left);
                    pathQueue.offer(new StringBuffer(path).append("->").append(node.left.val).toString());
                }

                if (node.right != null) {
                    nodeQueue.offer(node.right);
                    pathQueue.offer(new StringBuffer(path).append("->").append(node.right.val).toString());
                }
            }
        }
        return paths;
    }
}

知识点：无

总结：无