[LeetCode] 590. N-ary Tree Postorder Traversal (vs. LC145)_Node.js

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[LeetCode] 590. N-ary Tree Postorder Traversal (vs. LC145)

发布时间：2019-07-01 发布网站：脚本宝典

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590. N-ary Tree Postorder Traversal
Problem
Given an n-ary tree, return the postorder traversal of its nodes' values.
For example, given a 3-ary tree:

Return its postorder traversal as: [5,6,3,2,4,1].
Note: Recursive solution is trivial, could you do it iteratively?
Solution (Recursion)

      
      
      ArrayList<>();
        helper(root, res);
        return res;
    }
    private void helper(Node root, List res) {
        if (root == null) return;
        if (root.children != null) {
            for (Node child: root.children) {
                helper(child, res);
            }
        }
        res.add(root.val);
    }
}" title="" data-original-title="复制">
      
      
class Solution {
    public List<Integer> postorder(Node root) {
        List<Integer> res = new ArrayList<>();
        helper(root, res);
        return res;
    }
    private void helper(Node root, List<Integer> res) {
        if (root == null) return;
        if (root.children != null) {
            for (Node child: root.children) {
                helper(child, res);
            }
        }
        res.add(root.val);
    }
}
Solution (Iteration)
按顺序放入stack，正好level方面是从root到leaves，顺序方面是从最右到最左，因为stack是先入后出。这样最后reverse一下就是先左后右，先子后根。
巧妙。

      
      
      ArrayList<>();
        if (root == null) return res;
        Stack stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            Node node = stack.pop();
            res.add(node.val);
            for (Node child: node.children) {
                stack.push(child);
            }
        }
        Collections.reverse(res);
        return res;
    }
}" title="" data-original-title="复制">
      
      
class Solution {
    public List<Integer> postorder(Node root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        Stack<Node> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            Node node = stack.pop();
            res.add(node.val);
            for (Node child: node.children) {
                stack.push(child);
            }
        }
        Collections.reverse(res);
        return res;
    }
}
145. Binary Tree Postorder Traversal
Given a binary tree, return the postorder traversal of its nodes' values.
Example:
Input: [1,null,2,3]

      
      
      
      
      
   1
    
     2
    /
   3

Output: [3,2,1]
Follow up: Recursive solution is trivial, could you do it iteratively?
Solution (Iteration)

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

      
      
      
      
      

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