题意：根据中序和后序遍历构造二叉树

思路：

1. 把中序遍历的每一个数字和对应的index放到map中
2. DFS，遍历重构树
3. 每次DFS，传入中序和后序的开始和结束index，
4. 取后序index的末尾为跟节点
5. DFS获取其左右子节点
6. 返回跟节点

思想：树的先跟遍历

复杂度：时间O（n)，空间O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        int len = inorder.length;
        if(len == 0)
            return null;
        HashMap<Integer, Integer> map = new HashMap();
        for(int i=0;i<len;i++) {
            map.put(inorder[i], i);
        }
        
        return build(inorder, postorder, 0, len-1, 0, len-1, map);
    }
    TreeNode build(int[] inorder, int[] postorder, int is, int ie, int ps, int pe, HashMap<Integer, Integer> map) {
        if(is>ie || ps > pe || is<0||ps<0)
            return null;
        TreeNode cur = new TreeNode(postorder[pe]);
        int index = map.get(postorder[pe]);
        cur.left = build(inorder, postorder, is, index - 1, ps, ps + index - is-1, map);
        cur.right = build(inorder, postorder, index+1, ie, ps+index-is, pe-1, map);
        return cur;
    }
}