[数据结构与算法]leetcode145二叉树的后序遍历

给定一个二叉树，返回它的 后序 遍历。
示例:
输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [3,2,1]
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-postorder-traversal

递归

class Solution {
    List<Integer> list1 = new ArrayList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        if(root == null) return list1;
        return postorder(root); 
    }
    public List<Integer> postorder(TreeNode root){
        if(root.left != null){
            postorder(root.left);
        }
        if(root.right != null){
            postorder(root.right);
        }
        list1.add(root.val);
        return list1;
    }
}

迭代1 使用LinkedList

class Solution {
    LinkedList<Integer> list1 = new LinkedList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        if(root == null) return list1;
        return postorder(root); 
    }
    public List<Integer> postorder(TreeNode root){
        LinkedList<TreeNode> l1 = new LinkedList<>();
        l1.add(root);
        while(!l1.isEmpty()){
            TreeNode n1 = l1.pollLast();
            list1.addFirst(n1.val);   //addFirst---LinkedList中的方法
            if(n1.left != null) l1.add(n1.left);
            if(n1.right != null) l1.add(n1.right);         
        }
        return list1;
    }
}

迭代 stack

class Solution {
    LinkedList<Integer> list1 = new LinkedList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        if(root == null) return list1;
        return postorder(root); 
    }
    public List<Integer> postorder(TreeNode root){
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode n1 = stack.pop();
            if (n1 == null) continue;
            list1.addFirst(n1.val);
            stack.push(n1.left);
            stack.push(n1.right);
        }
        return list1;
    }
}