LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树 C++

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

    3
   /   9  20
    /     15   7

中序、后序遍历得到二叉树，可以知道每一次新数组的最后一个数为当时子树的根节点，每次根据中序遍历的根节点的左右两边确定左右子树，再对应后序的左右子树，不停递归得到根节点，可以建立二叉树。每次由循环得到根节点在中序数组中坐标i

由中序遍历知：每一次inorder的左子树范围[ileft，i-1],右子树范围[i+1,iright]

LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树 C++

原文：https://www.cnblogs.com/hhhhan1025/p/10713560.html