[LeetCode] Binary Tree Paths

Binary Tree Paths

Recursion

Time Complexity
O(N)

Space Complexity
O(logn)

思路

树的前序遍历，递归的终止条件的第一个root == null，另一个终止条件是当root是叶子节点的时候，就不用再往下递归了，同时加上自己的值，加入res

代码

public List<String> binaryTreePaths(TreeNode root) {
    // Write your code here
    List<String> res = new ArrayList<String>();
    if(root == null) return res;
    String path = "";
    helper(root, path, res);
    return res;
}

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