42-二叉树练习-LeetCode106从中序与后序遍历序列构造二叉树

题目

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]

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

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思路

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代码

/**
 * 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 {
    //优化：存储中序遍历的对应的值和索引
    Map map = new HashMap<>();

    public TreeNode buildTree(int[] inOrder, int[] postOrder) {
        //将后序遍历结果反转
        int[] preOrder = reverse(inOrder, postOrder);
        return buildTreeInternal(preOrder, inOrder, 0, inOrder.length);
    }

    int index = 0;

    public TreeNode buildTreeInternal(int[] preOrder, int[] inOrder, int left, int right) {
        if(left >= right) {
            return null;
        }
        if(index >= preOrder.length) {
            return null;
        }

        TreeNode root = new TreeNode(preOrder[index]);
        index++;
        int pos = map.get(root.val);
        root.right = buildTreeInternal(preOrder, inOrder, pos + 1, right);
        root.left = buildTreeInternal(preOrder, inOrder, left, pos);
        return root;
    }

    private int[] reverse(int[] inOrder, int[] postOrder) {
        int[] ret = new int[postOrder.length];
        for (int i = 0; i < ret.length; i++) {
            ret[i] = postOrder[ret.length - 1 - i]; //postOrder[ret.length - 1]最后一个元素
        }
        for (int i = 0; i < inOrder.length; i++) {
            map.put(inOrder[i], i); //将中序遍历序列存入map中
        }
        return ret; //后序遍历序列反转后得到的镜像前序遍历
    }
}