LeetCode | Construct Binary Tree from Preorder and Inorder Traversal

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

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

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

 //从先序和中序遍历恢复树，主要思想是利用递归，抽象算法如下：
 // TreeNode root = preorder[0];
 // root.left  = buildTree(preorder[leftsubtree], inorder[leftsubtree]);
 // root.right = buildTree(preorder[rightsubtree], inorder[rightsubtree]);
 // 先序遍历数组的首一定是root，然后根据root.val去中序遍历中定位root的位置
 // 算法的关键就在于每次递归时在先序与中序遍历中定位出leftsubtree与rightsubtree

public class Solution {
    public int findRoot(int[] inorder, int key){      //在中序遍历中定位root的位置下标
        if(inorder.length==0) return -1;
        int index = -1;
        for(int i=0; i<inorder.length; i++){
            if(inorder[i]==key){
                index = i;break;
            }
        }
        return index;
    }
    
    //实现重建树的逻辑，[prestart,preend]表示先序遍历的位置，[instart,inend]表示中序遍历的位置
    public TreeNode buildTree(int[] preorder, int[] inorder, int prestart, int preend, int instart, int inend){
        if(preorder.length==0 || inorder.length==0)  return null;
        if(instart>inend)  return null;
         
        TreeNode root = new TreeNode(preorder[prestart]);
        int index = findRoot(inorder, root.val);   //index表示root在中序遍历inorder中的位置
        
        //下面两行要自己画图才能看出来，        preorder[leftsubtree]= preorder[prestart+1至prestart+1+index-1-instart]
        // 以在先序、中序遍历中定位左子树为例：  inorder[leftsubtree]= inorder[instart至index-1]
        root.left = buildTree(preorder, inorder, prestart+1, prestart+1+index-1-instart, instart, index-1);
        root.right = buildTree(preorder, inorder, prestart+1+index-1-instart+1, preend, index+1, inend);
        return root;
    }
    
    //******leetcode主函数******
    public TreeNode buildTree(int[] preorder, int[] inorder) {   
        return buildTree(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1);    
    }
}