构建二叉树-leetcode106

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 

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]

提示:

• 1 <= inorder.length <= 3000
• postorder.length == inorder.length
• -3000 <= inorder[i], postorder[i] <= 3000
• inorder 和 postorder 都由 不同 的值组成
• postorder 中每一个值都在 inorder 中
• inorder 保证是树的中序遍历
• postorder 保证是树的后序遍历

//leetcode submit region begin(Prohibit modification and deletion)
/**
 * 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 {
    // 存储 inorder 中值到索引的映射
    HashMap<Integer, Integer> valToIndex = new HashMap<>();

    TreeNode buildTree(int[] inorder, int[] postorder) {
        for (int i = 0; i < inorder.length; i++) {
            valToIndex.put(inorder[i], i);
        }
        return build(inorder, 0, inorder.length - 1,
                postorder, 0, postorder.length - 1);
    }

    /* 
        build 函数的定义：
        后序遍历数组为 postorder[postStart..postEnd]，
        中序遍历数组为 inorder[inStart..inEnd]，
        构造二叉树，返回该二叉树的根节点 
    */
    TreeNode build(int[] inorder, int inStart, int inEnd,
                   int[] postorder, int postStart, int postEnd) {
        // root 节点对应的值就是后序遍历数组的最后一个元素
        int rootVal = postorder[postEnd];
        // rootVal 在中序遍历数组中的索引
        int index = valToIndex.get(rootVal);

        TreeNode root = new TreeNode(rootVal);
        // 递归构造左右子树
        root.left = build(preorder, ?, ?,
        inorder, ?, ?);

        root.right = build(preorder, ?, ?,
        inorder, ?, ?);
        return root;
    }

}
//leetcode submit region end(Prohibit modification and deletion)