Construct Binary Tree from Preorder and Inorder Traversal

来源：https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length == 0) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[0]);
        int rootIndex = 0;
        while(rootIndex < inorder.length) {
            if(inorder[rootIndex] == preorder[0]) {
                break;
            }
            rootIndex++;
        }
        int[] preLeftSubTree = Arrays.copyOfRange(preorder, 1, rootIndex+1);
        int[] preRightSubTree = Arrays.copyOfRange(preorder, rootIndex+1, preorder.length);
        int[] inLeftSubTree = Arrays.copyOfRange(inorder, 0, rootIndex);
        int[] inRightSubTree = Arrays.copyOfRange(inorder, rootIndex+1, inorder.length);
        root.left = buildTree(preLeftSubTree, inLeftSubTree);
        root.right = buildTree(preRightSubTree, inRightSubTree);
        return root;
    }
}// 36 ms，待优化，去掉数组拷贝

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):
        if len(pre) == 0:
            return None
        if len(pre) == 1:
            return TreeNode(pre[0])
        head = TreeNode(pre[0])
        head.left = self.reConstructBinaryTree(pre[1:tin.index(pre[0])+1], tin[:tin.index(pre[0])])
        head.right = self.reConstructBinaryTree(pre[tin.index(pre[0])+1:], tin[tin.index(pre[0])+1:])
        return head