589. N 叉树的前序遍历

java1：stack栈：没看懂

class Solution {public List<Integer> preorder(Node root) {List<Integer> res = new ArrayList<Integer>();if (root == null) {return res;}Map<Node, Integer> map = new HashMap<Node, Integer>();Deque<Node> stack = new ArrayDeque<Node>();Node node = root;while (!stack.isEmpty() || node != null) {while (node != null) {res.add(node.val);  // 添加入resstack.addFirst(node);List<Node> children = node.children;if (children != null && children.size() > 0) {map.put(node, 0);node = children.get(0);  // 一直往左侧延伸，所以一直get(0)} else {node = null;}}node = stack.peekFirst();int index = map.getOrDefault(node, -1) + 1;  // 当前节点map-get一次，遍历一个分支！List<Node> children = node.children;  // 延伸到最左下边，然后check右侧的节点if (children != null && children.size() > index) {  // node移至右侧一个节点，然后再循环往左下延伸！map.put(node, index);node = children.get(index);} else {  // 子节点不存在 || children使用结束，那么当前node就应该删除、并且node==null，不进入下一个while，进入右侧下一个节点stack.removeFirst();map.remove(node);node = null;  // 不进入下一次while}}return res;}
}

java2：DFS

class Solution {List<Integer> list = new ArrayList<>();private void dfs(Node root) {if (root == null) {return;}list.add(root.val);for (Node child : root.children) {dfs(child);}}public List<Integer> preorder(Node root) {dfs(root);return list;}
}