package 二叉树.BT;
import java.awt.HeadlessException;
import java.util.HashMap;
import java.util.Map;
/**
 * https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
 * 
 * @author Huangyujun
 *
 */
public class _105_从前序与中序遍历序列构造二叉树 {
    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;
        }
    }
    /**
     * preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
     * 前序特点：第一个便是根结点（知道根比较快），中序：左 根 右
     * 前序提供的根结点， 中序不断地划分成左子树、右子树
     * @param preorder
     * @param inorder
     * @return
     */
    //定义一个 map，键值对，【值， 位置】
    Map<Integer, Integer> map = new HashMap<>();
    int[] inorder;
    int[] preorder;
    int pre_id = 0;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        this.preorder = preorder;
        this.inorder = inorder;
        int n = inorder.length;
        for(int i = 0; i < n; i++) {
            map.put(inorder[i], i);
        }
        return helper(0, n - 1);
    }
    //构建的helper（表示一棵树的范围）
    public TreeNode helper(int left, int right) {
        if(left > right)    return null;    
         int rootVal = preorder[pre_id];
        //创建根结点
        TreeNode root = new TreeNode(rootVal);
        //下标++ ，方便下次拿到左子树的根结点
        pre_id++;
        //获取根在中序的位置，将分成两个树
        int index = map.get(rootVal);
        //左子树
        root.left = helper(left, index - 1);
        //右子树
        root.right = helper(index + 1, right);
        return root;
    }
}