Go Java算法之二叉树的所有路径示例详解

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);
            }
        }
    }
}

func binaryTreePaths(root *TreeNode) []string {
    paths := []string{}
    if root == nil {
        return paths
    }
    nodeQueue := []*TreeNode{}
    pathQueue := []string{}
    nodeQueue = append(nodeQueue, root)
    pathQueue = append(pathQueue, strconv.Itoa(root.Val))
    for i := 0; i < len(nodeQueue); i++ {
        node, path := nodeQueue[i], pathQueue[i]
        if node.Left == nil && node.Right == nil {
            paths = append(paths, path)
            continue
        }
        if node.Left != nil {
            nodeQueue = append(nodeQueue, node.Left)
            pathQueue = append(pathQueue, path + "->" + strconv.Itoa(node.Left.Val))
        }
        if node.Right != nil {
            nodeQueue = append(nodeQueue, node.Right)
            pathQueue = append(pathQueue, path + "->" + strconv.Itoa(node.Right.Val))
        }
    }
    return paths
}