Construct Binary Tree from Inorder and Postorder Traversal:由中序后序遍历构造二叉树

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

思路：递归构造，因为后序遍历中最后一个节点是根节点，因此类似于先序遍历，根据后序根节点将中序分为左右两部分，递归构建左右子树。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
public TreeNode build(int[] inorder,int inLeft,int inRight,int[] postorder,int postLeft, int postRight){
        if(inLeft > inRight ||postLeft > postRight) return null;
        TreeNode n = new TreeNode(postorder[postRight]);
        int roorIndex = -1;
        for(int i = inLeft; i <= inRight; i++){
            if(inorder[i] == postorder[postRight]){
                roorIndex = i;
                break;
            }
        }
        int lLen = roorIndex - inLeft;
        int RLen = inRight - roorIndex;
        n.right = build(inorder,roorIndex + 1,inRight,postorder,postRight - RLen,postRight - 1);
        n.left = build(inorder,inLeft,roorIndex - 1,postorder,postLeft,postLeft + lLen - 1);
        return n;
    }
    
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return build(inorder,0,inorder.length - 1,postorder,0,postorder.length - 1);
    }
}