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剑指 Offer 07. 重建二叉树
阅读量:85 次
发布时间:2019-02-26

本文共 1183 字,大约阅读时间需要 3 分钟。

问题描述

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

解决思路:

树的相关问题,一般可以考虑使用递归的方法解决,递归树。

class Solution {       Map
map = new HashMap<>(); int[] preorder; public TreeNode buildTree(int[] preorder, int[] inorder) { //重建二叉树,使用分治的方法 this.preorder = preorder; //Map缓存 for(int i = 0; i < inorder.length; i++) { map.put(inorder[i], i); } //回溯 (1)打断条件 (2)逻辑处理 (3)下探 (4)回溯 return trackck(0, 0, preorder.length - 1); } public TreeNode trackck(int rootPreIndex, int inorderLeft, int inorderRight) { if(inorderLeft > inorderRight) return null; //获取中序遍历根节点,用于进行左右子树的划分 TreeNode root = new TreeNode(preorder[rootPreIndex]); int rootInIndex = map.get(preorder[rootPreIndex]); root.left = trackck(rootPreIndex + 1, inorderLeft, rootInIndex - 1); root.right = trackck(rootPreIndex + rootInIndex - inorderLeft + 1, rootInIndex + 1, inorderRight); return root; }}

来源:力扣(LeetCode)

链接:https://leetcode-cn.com/problems/zhong-jian-er-cha-shu-lcof/

转载地址:http://wxsu.baihongyu.com/

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