根据二叉树的前序遍历和中序遍历的结果，重建二叉树

题目描述

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。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。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

先定义一个二叉树类TreeNode

package com.jason.algorithm;
 public class TreeNode {
      int val;
      TreeNode left;
      TreeNode right;
      TreeNode(int x) { val = x; }
  }

具体实现方法如下：

package com.jason.algorithm;

import java.util.HashMap;

public class Solution {

	public static void main(String[] args) {
		int pre[] = { 1, 2, 4, 7, 3, 5, 6, 8 };
		int in[] = { 4, 7, 2, 1, 5, 3, 8, 6 };
		Solution solution = new Solution();
		solution.reConstructBinaryTree(pre, in);

	}

	public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
		if (pre == null || in == null) {
			return null;
		}
		HashMap<Integer, Integer> hashMap = new HashMap<Integer, Integer>();
		for (int i = 0; i < in.length; i++) {
			hashMap.put(in[i], i);
		}
		return preIn(pre, 0, pre.length - 1, in, 0, in.length - 1, hashMap);
	}

	// {1,2,4,7,3,5,6,8} pre 根左右
	// {4,7,2,1,5,3,8,6} in 左根右

	public TreeNode preIn(int[] p, int pi, int pj, int[] n, int ni, int nj,
			HashMap<Integer, Integer> map) {
		if (pi > pj) {
			return null;
		}
		TreeNode head = new TreeNode(p[pi]);
		int index = map.get(p[pi]);
		// 根据前序遍历知道根节点，根据中序遍历知道根左边的为左子树，右边的为右子树
		head.left = preIn(p, pi + 1, pi + index - ni, n, ni, index - 1, map);
		head.right = preIn(p, pi + index - ni + 1, pj, n, index + 1, nj, map);
		System.out
				.println("head =" + (head != null ? head.val : " ")
						+ " head.left ="
						+ (head.left != null ? head.left.val : " ")
						+ " head.right ="
						+ (head.right != null ? head.right.val : " "));
		return head;
	}
}