剑指 Offer 07. 重建二叉树

/**
 *	剑指 Offer 07. 重建二叉树
 * @author wsq
 * @date 2020/09/11
	 输入某二叉树的前序遍历和中序遍历的结果，请重建该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。

	例如，给出
	前序遍历 preorder = [3,9,20,15,7]
	中序遍历 inorder = [9,3,15,20,7]
	返回如下的二叉树：
	    3
	   / \
	  9  20
	    /  \
	   15   7
	
	链接：https://leetcode-cn.com/problems/zhong-jian-er-cha-shu-lcof
 */
package notsubmit;

public class RebuildTree {
	/**
	 *	采用递归的方式实现，由于每次递归都需要对原有的数组按照要求复制一部分，浪费了很多的时间和空间
	 * @param preorder
	 * @param inorder
	 * @return
	 */
	public TreeNode buildTree(int[] preorder, int[] inorder) {
		int val = preorder[0];
		int leftNum = getIndex(inorder, val);
		int rightNum = preorder.length - leftNum - 1;
		
		TreeNode node = new TreeNode(val);
	
		if(leftNum != 0) {
			// 需要优化，改为数组索引的形式
			int[] leftPreorder = new int[leftNum];
			System.arraycopy(preorder, 1, leftPreorder, 0, leftNum);
			int[] leftInorder = new int[leftNum];
			System.arraycopy(inorder, 0, leftInorder, 0, leftNum);
			node.left = buildTree(leftPreorder, leftInorder);
		}
		if(rightNum != 0) {			
			int[] rightPreorder = new int[rightNum];
			System.arraycopy(preorder, leftNum + 1, rightPreorder, 0, rightNum);
			int[] rightInorder = new int[rightNum];
			System.arraycopy(inorder, leftNum + 1, rightInorder, 0, rightNum);
			node.right = buildTree(rightPreorder, rightInorder);
		}
		
		return node;
    }
	
	public int getIndex(int[] arr, int val) {
		int ans = 0;
		for(int i = 0; i < arr.length; ++i) {
			if(arr[i] == val) {
				ans = i;
				break;
			}
		}
		return ans;
	}
	
	public static void main(String[] args) {
		int[] preorder = {3,9,20,15,7};
		int[] inorder = {9,3,15,20,7};
		RebuildTree rt = new RebuildTree();
		TreeNode ndoe = rt.buildTree(preorder, inorder);
	}
}