根据后序遍历和中序遍历构建二叉树

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
        //从后续遍历的数组的最后一个元素开始找。  先构建右树 再构建左树。
     public int preIndex = 0;
    public TreeNode buildTreeChild(int[] preorder, int[] inorder,int inbegin,int inend) {
        
        if(inbegin > inend) {
            return null;
        }
        
        TreeNode root = new TreeNode(preorder[preIndex]);
        
        //找preorder[preIndex] 在中序遍历的数组当中的下标 记为rootIndex;
        int rootIndex = findIndexOfInorder(inorder,preorder[preIndex],inbegin,inend);
       
        preIndex--;
        
         root.right = buildTreeChild(preorder,inorder,rootIndex+1,inend);
        
        root.left = buildTreeChild(preorder,inorder,inbegin,rootIndex-1);
       
       
        return root;
        
    }
    //找到返回下标  找不到返回-1
    public int findIndexOfInorder(int[] inorder,int val,int inbegin,int inend) {
        //找
        for(int i = inbegin;i <= inend;i++) {
            if(inorder[i] == val) {
                return i;
            }
        }
        return -1;
    }
    
    public TreeNode buildTree(int[] inorder,int[] preorder) {
        if(preorder==null || inorder==null){
            return null;
        }
        preIndex = preorder.length-1;
        return buildTreeChild(preorder,inorder,0,inorder.length-1);
    }
}