Construct Binary Tree from Preorder and Inorder Traversal Java

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本文详细介绍了如何使用哈希表和递归算法构建二叉树，从预序遍历和中序遍历出发，通过解析根节点和左右子树的位置关系，实现树的构建。

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

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

Key to Solve: Recursion + HashMap(Key: Inorder's Node, Value: position)
Preorder: we can see the top Root is 1st element of Preorder, and so on
Inorder: the elements before the root are belong to left subtree
after the root are belong to right subtree
The split left and right subtree base on value of Preorder
and position of inorder

public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
  if(preorder.length==0 || inorder.length==0) return null;
        HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();
        for(int i=0;i<inorder.length;i++){

            map.put(inorder[i],i);
        }
        return helperRecur(preorder,inorder,0,preorder.length-1,0,inorder.length-1,map);
    }
    private TreeNode helperRecur(int[] preorder, int[] inorder, int preL, int preR,
                               int inL, int inR, HashMap<Integer,Integer> map){
        if(preL>preR || inL>inR){
            return  null;
        }
        TreeNode root=new TreeNode(preorder[preL]);
        int rootIndex=map.get(root.val);
        int distance=rootIndex-inL;
        root.left=helperRecur(preorder,inorder,preL+1,preL+distance,inL,rootIndex-1,map);
        root.right=helperRecur(preorder,inorder,preL+distance+1,preR,rootIndex+1,inR,map);

        return root;
    }
}