[LeetCode]Construct Binary Tree from Inorder and Postorder Traversal

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

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

基本思想和知道前序与中序的思想一样，中序的某一节点的左节点一定是该节点的左子树，而后序遍历的某一节点的左节点一定是该节点的右子树，然后递归

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {public TreeNode buildTree(int[] inorder, int[] postorder) {    return buildTree(inorder,postorder,0,inorder.length-1,0,postorder.length-1);}private TreeNode buildTree(int []inorder, int []postorder,int inst,int inend,int postst, int postend){if(inst>inend||postst>postend||inorder.length<1){return null;}TreeNode tn =new TreeNode(postorder[postend]);        int index = Arrays.binarySearch(inorder, inst, inend+1, postorder[postend]);tn.left = buildTree(inorder,postorder,inst,inst+index-1,postst,postst+index-inst-1);tn.right = buildTree(inorder,postorder,index+1,inend,postst+index-inst,postend-1);return tn;}}