[LeetCode] 145. Binary Tree Postorder Traversal

Given the root of a binary tree, return the postorder traversal of its nodes' values.

Example 1:

Input: root = [1,null,2,3]
Output: [3,2,1]

Example 2:
Input: root = []
Output: []

Example 3:
Input: root = [1]
Output: [1]

Constraints:
The number of the nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100

Follow up: Recursive solution is trivial, could you do it iteratively?

给你一棵二叉树的根节点 root ，返回其节点值的 后序遍历 。

思路

依然是迭代和递归两种做法，两种做法的时间复杂度均是O(n)，空间复杂度均是O(h)。
递归没什么好讲的，直接上代码。

复杂度

时间O(n)
空间O(h)

代码

迭代

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        if (root == null) {
			return res;
		}
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.addFirst(cur.val);
            if (cur.left != null) {
				stack.push(cur.left);
			}
            if (cur.right != null) {
				stack.push(cur.right);
			}
        }
        return res;
    }
}

递归

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
			return res;
		}
        helper(res, root);
        return res;
    }

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