剑指offer-题6：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。

实验平台：牛客网

和二叉树相关的遍历方式如下：





java:

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
TreeNode root = new TreeNode(pre[0]);
root.left = null;
root.right = null;
int rootIndex = 0; // 根节点在中序遍历的位置
while (rootIndex < in.length && root.val != in[rootIndex]) {
rootIndex++;
}

int rightLength = in.length - rootIndex - 1; // 根节点右子树节点个数

if (rootIndex > 0) { // 构造左子树
int[] inTemp = new int[rootIndex];
int[] preTemp = new int[rootIndex];
for (int i = 0; i < rootIndex; i++) {
inTemp[i] = in[i];
}
for (int j = 1; j <= rootIndex; j++) {
preTemp[j - 1] = pre[j];
}

root.left = reConstructBinaryTree(preTemp, inTemp);
}
if (rightLength > 0) { // 构造右子树
int[] inTemp = new int[rightLength];
int[] preTemp = new int[rightLength];
for (int i = 0; i < rightLength; i++) {
inTemp[i] = in[rootIndex + 1 + i];
}
for (int j = 1; j <= rightLength; j++) {
preTemp[j - 1] = pre[rootIndex + j];
}

root.right = reConstructBinaryTree(preTemp, inTemp);
}
return root;
}
}

python:

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
# 返回构造的TreeNode根节点
def reConstructBinaryTree(self, pre, tin):
# write code here
node = TreeNode(pre[0])
root_index = tin.index(pre[0])
right_length = len(tin) - root_index - 1
if root_index > 0:
node.left = self.reConstructBinaryTree(pre[1: root_index + 1], tin[0:root_index])  # 构造左子树
if right_length > 0:
node.right = self.reConstructBinaryTree(pre[root_index + 1:], tin[root_index + 1:])  # 构造右子树
return node