問題描述

根據一棵樹的中序遍歷與後序遍歷構造二叉樹。

注意:
你可以假設樹中沒有重複的元素。

例如，給出

中序遍歷 inorder =[9,3,15,20,7]
後序遍歷 postorder = [9,15,7,20,3]
返回如下的二叉樹：

3
/ \
9 20
/ \
15 7

來源：力扣（LeetCode）
連結：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);
    }
}