LeetCode | Construct Binary Tree from Inorder and Postorder Traversal

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

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

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

 //从中序与后序遍历构建树，思想与上一题一样
 //唯一区别是后序遍历的尾为root，而先序遍历的首为root
 public class Solution {
     public int findRoot(int[] inorder, int key){     //在中序遍历中定位root
         if(inorder.length==0) return -1;
         int index = -1;
         for(int i=0; i<inorder.length; i++){
             if(inorder[i]==key){
                 index = i; break;
             }
         }
         return index;
     }
     
     //通过递归实现构建树的主逻辑
     public TreeNode buildTree(int[] inorder, int[] postorder, int instart, int inend, int postart, int poend){
         if(inorder.length==0 || postorder.length==0) return null;
         if(instart > inend) return null;
         TreeNode root = new TreeNode(postorder[poend]);
         int index = findRoot(inorder, root.val);
         root.left = buildTree(inorder, postorder, instart, index-1, postart, postart+index-1-instart);
         root.right = buildTree(inorder, postorder, index+1, inend, postart+index-1-instart+1, poend-1);
         return root;
     }
     
     //****leetcode主函数****
     public TreeNode buildTree(int[] inorder, int[] postorder) {
         return buildTree(inorder, postorder, 0, inorder.length-1, 0, postorder.length-1);
     }
 }