【LeetCode】145. 二叉树的后序遍历

题目描述：

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]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-postorder-traversal
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递归写法

/**
 * 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) {
    
    
        List<Integer> list = new ArrayList<>();
        postorderTraversal(root, list);
        return list;
    }
    public void postorderTraversal(TreeNode root, List<Integer> list) {
    
    
        if (root == null) return;
        postorderTraversal(root.left, list);
        postorderTraversal(root.right, list);
        list.add(root.val);
    }
}

迭代写法

/**
 * 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) {
    
    
        List<Integer> list = new ArrayList<>();
        if (root == null) return list;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode prev = null;
        while (!stack.isEmpty() || root != null) {
    
    
            while (root != null) {
    
    
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            if (root.right == null || root.right == prev) {
    
    
                list.add(root.val);
                prev = root;
                root = null;
            } else {
    
    
                stack.push(root);
                root = root.right;
            }
        }
        return list;
    }
}