给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。

示例 1:

输入：inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
输出：[3,9,20,null,null,15,7]

示例 2:

输入：inorder = [-1], postorder = [-1]
输出：[-1]

解法一

常规解法，利用递归，传递左子树和右子树的数组范围即可。

public TreeNode buildTree(int[] inorder, int[] postorder) {HashMap<Integer, Integer> map = new HashMap<>();for (int i = 0; i < inorder.length; i++) {map.put(inorder[i], i);}return buildTreeHelper(inorder, 0, inorder.length, postorder, 0, postorder.length, map);
}private TreeNode buildTreeHelper(int[] inorder, int i_start, int i_end, int[] postorder, int p_start, int p_end,HashMap<Integer, Integer> map) {if (p_start == p_end) {return null;}int root_val = postorder[p_end - 1];TreeNode root = new TreeNode(root_val);int i_root_index = map.get(root_val);int leftNum = i_root_index - i_start;root.left = buildTreeHelper(inorder, i_start, i_root_index, postorder, p_start, p_start + leftNum, map);root.right = buildTreeHelper(inorder, i_root_index + 1, i_end, postorder, p_start + leftNum, p_end - 1,map);return root;
}

解法二 stop 值

这里的话，之前说了，递归的话得先构造右子树再构造左子树，此外各种指针，也应该从末尾向零走。

视线从右往左看。

    3/ \9  20/  \15   7s 初始化一个树中所有的数字都不会相等的数，所以代码中用了一个 long 来表示
<------------------
中序9, 3, 15, 20, 7
^               ^
s               i后序
9, 15, 7, 20, 3^  p
<-------------------

p 和 i 都从右往左进行遍历，所以 p 开始产生的每次都是右子树的根节点。之前代码里的++要相应的改成--。

int post;
int in; 
public TreeNode buildTree(int[] inorder, int[] postorder) {post = postorder.length - 1;in = inorder.length - 1;return buildTreeHelper(inorder, postorder, (long) Integer.MIN_VALUE - 1);
}private TreeNode buildTreeHelper(int[] inorder, int[] postorder, long stop) {if (post == -1) {return null;}if (inorder[in] == stop) {in--;return null;}int root_val = postorder[post--];TreeNode root = new TreeNode(root_val);root.right = buildTreeHelper(inorder, postorder, root_val);root.left = buildTreeHelper(inorder, postorder, stop);return root;
}

解法三 栈

之前解法是构造左子树、左子树、左子树，出现相等，构造一颗右子树。这里相应的要改成构造右子树、右子树、右子树，出现相等，构造一颗左子树。和解法二一样，两个指针的话也是从末尾到头部进行。

public TreeNode buildTree(int[] inorder, int[] postorder) {if (postorder.length == 0) {return null;}Stack<TreeNode> roots = new Stack<TreeNode>();int post = postorder.length - 1;int in = inorder.length - 1;TreeNode curRoot = new TreeNode(postorder[post]);TreeNode root = curRoot;roots.push(curRoot);post--;while (post >=  0) {if (curRoot.val == inorder[in]) {while (!roots.isEmpty() && roots.peek().val == inorder[in]) {curRoot = roots.peek();roots.pop();in--;}curRoot.left = new TreeNode(postorder[post]);curRoot = curRoot.left;roots.push(curRoot);post--;} else {curRoot.right = new TreeNode(postorder[post]);curRoot = curRoot.right;roots.push(curRoot);post--;}}return root;
}