Construct Binary Tree from Inorder and Postorder Traversal, inorderpostorder

Construct Binary Tree from Inorder and Postorder Traversal, inorderpostorder

The original question is as follows;

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

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

The meaning is simple: generate a tree based on the ordinal traversal and the ordinal traversal.

The idea is simple: determine the root node based on the Post-order traversal sequence, locate the root node in the middle-order traversal, and divide the original sequence into two parts, recursion solves this problem by solving the Left and Right Parts.

The Java code is as follows:

Public class ConstructBinaryTreefromInorderandPostorderTraversal {/*** @ Construct Binary Tree from Inorder and Postorder Traversal * link: https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/ * idea: recursion */class TreeNode {int val; TreeNode left; TreeNode right; public TreeNode (int x) {val = x ;}} class Solution {public TreeNode buildTree (int [] inorder, int [] postorder) {if (inorder. length = 0) {return null;} TreeNode node = createTree (inorder, 0, inorder. length-1, postorder, 0, postorder. length-1); return node;} public TreeNode createTree (int [] inorder, int inbegin, int inend, int [] postorder, int postbegin, int postend) {if (inbegin> inend) {return null;} int root = postorder [postend]; int index = 0; for (int I = inbegin; I <= inend; I ++) {if (root = inorder [I]) {index = I; break;} int len = index-inbegin; TreeNode left = createTree (inorder, inbegin, index-1, postorder, postbegin, postbegin + len-1); TreeNode right = createTree (inorder, index + 1, inend, postorder, postbegin + len, postend-1 ); treeNode node = new TreeNode (root); node. left = left; node. right = right; return node ;}}}