[抄题]:
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.For example, given
preorder = [3,9,20,15,7]inorder = [9,3,15,20,7]
Return the following binary tree:
3 / \ 9 20 / \ 15 7
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
if (preStart > preEnd || inStart > inEnd) 退出条件是>,因为等于的时候还可以继续新建节点
[思维问题]:
不知道怎么dc:参数就是数组的index就行了,分成start1 end1 start2 end2,多开几个变量就行了
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
通过hashmap记录下pre的中节点在in中的位置,然后左右dc
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- HashMap<Integer, Integer> map 函数中已经建立好了的map不需要加括号,新建map才需要
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
不知道怎么dc:参数就是数组的index就行了,分成start1 end1 start2 end2,多开几个变量就行了
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
/** * 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) { //corner case if (preorder == null && inorder == null) return null; //initialization : put all the inorder into map HashMapmap = new HashMap (); for (int i = 0; i < inorder.length; i++) map.put(inorder[i], i); //return return buildTreeHelper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, map); } public TreeNode buildTreeHelper(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd, HashMap map) { //exit if the bounds exceeds if (preStart > preEnd || inStart > inEnd) return null; //build a new root TreeNode root = new TreeNode(preorder[preStart]); int inIdx = map.get(root.val); int numsLeft = inIdx - inStart; //divid and conquer to form left and right root.left = buildTreeHelper(preorder, preStart + 1, preEnd, inorder, inStart, inIdx - 1, map); root.right = buildTreeHelper(preorder, preStart + numsLeft + 1, preEnd, inorder, inIdx + 1, inEnd, map); return root; }}