106. 从中序与后序遍历序列构造二叉树-中等难度

问题描述

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。

例如，给出

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]
返回如下的二叉树：

3
/ \
9 20
/ \
15 7

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal

解答

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int[] postorder2 = null;
    int[] inorder2 = null;
    Map<Integer,Integer> map = null;
    int len = 0;
    public TreeNode recursive(int inorder_left,int inorder_right,int postorder_left,int postorder_right){
        if(inorder_right-inorder_left<=0)return null;
        if(inorder_right-inorder_left==1)return new TreeNode(inorder2[inorder_left]);
        int position = map.get(postorder2[postorder_right-1]);
        TreeNode r = null;
        if(position+1<len)
        r = recursive(position+1,inorder_right,postorder_right-(inorder_right-position),postorder_right-1);
        TreeNode l = recursive(inorder_left,position,postorder_left,postorder_left+position-inorder_left);
        return new TreeNode(postorder2[postorder_right-1],l,r);
    }
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        len = inorder.length;
        if(len==0)return null;
        postorder2 = postorder;
        inorder2 = inorder;
        map = new HashMap<Integer,Integer>();
        for(int i=0;i<len;i++)
            map.put(inorder2[i],i);
        return recursive(0,len,0,len);
    }
}