[LeetCode] 589. N-ary Tree Preorder Traversal (vs. LC144)_Node.js

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[LeetCode] 589. N-ary Tree Preorder Traversal (vs. LC144)

发布时间：2019-06-07 发布网站：脚本宝典

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589. N-ary Tree Preorder Traversal
Given an n-ary tree, return the preorder traversal of its nodes' values.
For example, given a 3-ary tree:

Return its preorder traversal as: [1,3,5,6,2,4].
Note: Recursive solution is trivial, could you do it iteratively?
Solution (Iteration)
Using stack, push the child from the end of list

      
      
      ArrayList<>();
        if (root == null) return res;
        Stack stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            Node cur = stack.pop();
            res.add(cur.val);
            for (int i = cur.children.size()-1; i >= 0; i--) stack.push(cur.children.get(i));
        }
        return res;
    }
}" title="" data-original-title="复制">
      
      

class Solution {
    public List<Integer> preorder(Node root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        Stack<Node> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            Node cur = stack.pop();
            res.add(cur.val);
            for (int i = cur.children.size()-1; i >= 0; i--) stack.push(cur.children.get(i));
        }
        return res;
    }
}
Solution (Recursion)

      
      
      ArrayList<>();
        helper(root, res);
        return res;
    }
    private void helper(Node root, List res) {
        if (root == null) return;
        res.add(root.val);
        for (Node node: root.children) {
            helper(node, res);
        }
    }
}" title="" data-original-title="复制">
      
      
/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List<Integer> preorder(Node root) {
        List<Integer> res = new ArrayList<>();
        helper(root, res);
        return res;
    }
    private void helper(Node root, List<Integer> res) {
        if (root == null) return;
        res.add(root.val);
        for (Node node: root.children) {
            helper(node, res);
        }
    }
}
144. Binary Tree Preorder Traversal
Given a binary tree, return the preorder traversal of its nodes' values.
Example:
Input: [1,null,2,3]

      
      
      
      
      
   1
    
     2
    /
   3

Output: [1,2,3]
Follow up: Recursive solution is trivial, could you do it iteratively?
Solution (Iteration)
Use stack, first push node.right, then push node.left

      
      
      
      
      
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            if (cur.right != null) stack.push(cur.right);
            if (cur.left != null) stack.push(cur.left);
        }
        return res;
    }
}

Solution (Recursion)

      
      
      
      
      
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        helper(root, res);
        return res;
    }
    private void helper(TreeNode root, List<Integer> res) {
        if (root == null) return;
        res.add(root.val);
        helper(root.left, res);
        helper(root.right, res);
    }
}