LeetCode 145. 二叉树的后序遍历

145. 二叉树的后序遍历

给你一棵二叉树的根节点?root?，返回其节点值的?后序遍历?。

示例 1：

输入：root = [1,null,2,3]
输出：[3,2,1]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

提示：

• 树中节点的数目在范围?[0, 100]?内
• -100 <= Node.val <= 100

进阶：递归算法很简单，你可以通过迭代算法完成吗？

解法思路：

1、递归

2、迭代

?法一：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        // Recursion
        // Time: O(n)
        // Space: O(n)
        List<Integer> res = new ArrayList<>();
        postorder(root, res);
        return res;
    }

    private void postorder(TreeNode root, List<Integer> res) {
        if (root == null) return;
        postorder(root.left, res);
        postorder(root.right, res);
        res.add(root.val);
    }
}

法二：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        // Iterator
        // Time: O(n)
        // Space: O(n)
        List<Integer> res = new ArrayList<>();
        if (root == null) return res;
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode prev = null;
        while (root != null || !stack.isEmpty()) {
            while (root != null) {
                stack.addLast(root);
                root = root.left;
            }
            root = stack.removeLast();
            if (root.right == null || root.right == prev) {
                res.add(root.val);
                prev = root;
                root = null;
            } else {
                stack.addLast(root);
                root = root.right;
            }
        }
        return res;
    }
}