binary-tree-postorder-traversal

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Algorithms

Given the <code>root</code> of a&nbsp;binary tree, return the postorder traversal of its nodes&#39; values.

&nbsp;
Example 1:

<div class="example-block">
Input: root = [1,null,2,3]

Output: [3,2,1]

Explanation:

<img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/screenshot-2024-08-29-202743.png" style="width: 200px; height: 264px;" />
</div>

Example 2:

<div class="example-block">
Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [4,6,7,5,2,9,8,3,1]

Explanation:

<img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/tree_2.png" style="width: 350px; height: 286px;" />
</div>

Example 3:

<div class="example-block">
Input: root = []

Output: []
</div>

Example 4:

<div class="example-block">
Input: root = [1]

Output: [1]
</div>

&nbsp;
Constraints:

<ul>
	<li>The number of the nodes in the tree is in the range <code>[0, 100]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

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

postorderTraversal

Binary Tree Postorder Traversal

Binary Tree Inorder Traversal

binary-tree-inorder-traversal

二叉树的中序遍历

N-ary Tree Postorder Traversal

n-ary-tree-postorder-traversal

N 叉树的后序遍历

Minimum Fuel Cost to Report to the Capital

minimum-fuel-cost-to-report-to-the-capital

到达首都的最少油耗

Stack

Tree

Depth-First Search

Binary Tree

给你一棵二叉树的根节点 <code>root</code> ，返回其节点值的 后序遍历 。

&nbsp;

示例 1：

<div class="example-block">
输入：root = [1,null,2,3]

输出：[3,2,1]

解释：

<img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/screenshot-2024-08-29-202743.png" style="width: 200px; height: 264px;" />
</div>

示例 2：

<div class="example-block">
输入：root = [1,2,3,4,5,null,8,null,null,6,7,9]

输出：[4,6,7,5,2,9,8,3,1]

解释：

<img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/tree_2.png" style="width: 350px; height: 286px;" />
</div>

示例 3：

<div class="example-block">
输入：root = []

输出：[]
</div>

示例 4：

<div class="example-block">
输入：root = [1]

输出：[1]
</div>

&nbsp;

提示：

<ul>
	<li>树中节点的数目在范围 <code>[0, 100]</code> 内</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

&nbsp;

进阶：递归算法很简单，你可以通过迭代算法完成吗？