刷题——通过前序中序遍历重建二叉树

/* * 题目描述Given preorder and inorder traversal of a tree, construct the binary tree.Note: You may assume that duplicates do not exist in the tree. */public class ConstructBinaryTreeFromPreorderAndInorderTraversal {public TreeNode buildTree(int[] preorder,int preStart,int preEnd, int[] inorder,int inStart,int inEnd) {if (preStart>preEnd || inStart>inEnd) {return null;}int rootVal = preorder[preStart];TreeNode rootNode = new TreeNode(rootVal);int i = inStart;for (; i <= inEnd; i++) {if (inorder[i]==rootVal) {//中序遍历中找到根break;}}rootNode.left = buildTree(preorder,preStart+1,preStart+i-inStart, inorder,inStart,i-1);rootNode.right = buildTree(preorder,preStart+i-inStart+1,preEnd, inorder,i+1,inEnd);return rootNode;}public TreeNode buildTree(int[] preorder, int[] inorder) {if (preorder==null || inorder==null ||preorder.length==0 || inorder.length==0) {return null;}        return buildTree(preorder,0,preorder.length-1, inorder,0,inorder.length-1);    }public static void main(String[] args) {// TODO Auto-generated method stub}}