construct-binary-tree-from-preorder-and-postorder-traversal

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Algorithms

Given two integer arrays, <code>preorder</code> and <code>postorder</code> where <code>preorder</code> is the preorder traversal of a binary tree of distinct values and <code>postorder</code> is the postorder traversal of the same tree, reconstruct and return the binary tree.

If there exist multiple answers, you can return any of them.

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/24/lc-prepost.jpg" style="width: 304px; height: 265px;" />
<pre>
Input: preorder = [1,2,4,5,3,6,7], postorder = [4,5,2,6,7,3,1]
Output: [1,2,3,4,5,6,7]
</pre>

Example 2:

<pre>
Input: preorder = [1], postorder = [1]
Output: [1]
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= preorder.length &lt;= 30</code></li>
	<li><code>1 &lt;= preorder[i] &lt;= preorder.length</code></li>
	<li>All the values of <code>preorder</code> are unique.</li>
	<li><code>postorder.length == preorder.length</code></li>
	<li><code>1 &lt;= postorder[i] &lt;= postorder.length</code></li>
	<li>All the values of <code>postorder</code> are unique.</li>
	<li>It is guaranteed that <code>preorder</code> and <code>postorder</code> are the preorder traversal and postorder traversal of the same binary tree.</li>
</ul>

constructFromPrePost

Construct Binary Tree from Preorder and Postorder Traversal

Tree

Array

Hash Table

Divide and Conquer

Binary Tree

给定两个整数数组，<code>preorder</code>&nbsp;和 <code>postorder</code> ，其中 <code>preorder</code> 是一个具有 无重复 值的二叉树的前序遍历，<code>postorder</code> 是同一棵树的后序遍历，重构并返回二叉树。

如果存在多个答案，您可以返回其中 任何 一个。

&nbsp;

示例 1：

<img src="https://assets.leetcode.com/uploads/2021/07/24/lc-prepost.jpg" style="height: 265px; width: 304px;" />

<pre>
输入：preorder = [1,2,4,5,3,6,7], postorder = [4,5,2,6,7,3,1]
输出：[1,2,3,4,5,6,7]
</pre>

示例 2:

<pre>
输入: preorder = [1], postorder = [1]
输出: [1]
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= preorder.length &lt;= 30</code></li>
	<li><code>1 &lt;= preorder[i] &lt;= preorder.length</code></li>
	<li><code>preorder</code>&nbsp;中所有值都 不同</li>
	<li><code>postorder.length == preorder.length</code></li>
	<li><code>1 &lt;= postorder[i] &lt;= postorder.length</code></li>
	<li><code>postorder</code>&nbsp;中所有值都 不同</li>
	<li>保证 <code>preorder</code>&nbsp;和 <code>postorder</code>&nbsp;是同一棵二叉树的前序遍历和后序遍历</li>
</ul>