106. 从中序与后序遍历序列构造二叉树

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]
返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public TreeNode buildTree2(int[] inorder, int[] postorder) {
        return build2(inorder, postorder, postorder.length - 1, 0, inorder.length - 1);
    }

    private TreeNode build2(int[] inorder, int[] postorder, int postEnd, int inSt, int inEnd) {
        //递归临界点
        if (postEnd < 0 || inSt > inEnd) {
            return null;
        }
        //后序遍历尾节点为根节点
        TreeNode rootNode = new TreeNode(postorder[postEnd]);
        //根节点在中序遍历的索引
        int rootIndex = 0;
        for (int i = inSt; i <= inEnd; i++) {
            if (inorder[i] == rootNode.val) {
                rootIndex = i;
                break;
            }
        }
        int rightLength = inEnd - rootIndex;
        int postEndLeft = postEnd - rightLength - 1;
        int postEndRight = postEnd - 1;
        rootNode.left = build2(inorder, postorder, postEndLeft, inSt, rootIndex - 1);
        rootNode.right = build2(inorder, postorder, postEndRight, rootIndex + 1, inEnd);
        return rootNode;
    }