【LeetCode】 106. Construct Binary Tree from Inorder and Postorder Traversal 从中序与后序遍历序列构造二叉树（JAVA）

【LeetCode】 106. Construct Binary Tree from Inorder and Postorder Traversal 从中序与后序遍历序列构造二叉树（Medium）（JAVA）

题目地址: https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

题目描述：

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

题目大意

根据一棵树的中序遍历与后序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

解题方法

和上一题一样，只是前序遍历和后序遍历的属性不同：【LeetCode】 105. Construct Binary Tree from Preorder and Inorder Traversal 从前序与中序遍历序列构造二叉树（JAVA）

1、后序遍历的根节点在最后一个；中序遍历的根节点在中间，找到根节点，就能分为左右子树
2、迭代更新完整的树

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < inorder.length; i++) {
            map.put(inorder[i], i);
        }
        return bH(inorder, postorder, map, 0, inorder.length - 1, 0, postorder.length - 1);
    }

    public TreeNode bH(int[] inorder, int[] postorder, Map<Integer, Integer> map, int iStart, int iEnd, int pStart, int pEnd) {
        if (iStart > iEnd || pStart > pEnd) return null;
        TreeNode root = new TreeNode(postorder[pEnd]);
        int mid = map.get(postorder[pEnd]);
        root.left = bH(inorder, postorder, map, iStart, mid - 1, pStart, pStart + mid - iStart - 1);
        root.right = bH(inorder, postorder, map, mid + 1, iEnd, pStart + mid - iStart, pEnd - 1);
        return root;
    }
}

执行用时 : 3 ms, 在所有 Java 提交中击败了 76.80% 的用户
内存消耗 : 40.4 MB, 在所有 Java 提交中击败了 61.90% 的用户