LeetCode Construct Binary Tree from Preorder and Inorder Traversal
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原题链接在这里:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
利用preorder 和 inorder来构造树。
e.g. preorder 1, 2, 4, 5, 3, 6, 7
inorder 4, 2, 5, 1, 6, 3, 7
先从preorder里去第一位,是1,1就是root;在inorder里找到1,inorder里1左面的[4, 2, 5]就都是 1的左子树inorder, inorder 里 1右面的[6, 3, 7]就都是1的右子树inorder。因为[4,2,5]长度为三,在preorder 1 后面的 三个数[2,4,5]就是就是对应左子树的preorder, 在后面的三个数[3,6,7]就是对应右子树的preorder.
重复上面的过程直到对应长度只有一个元素。
可以用HashMap来存储inorder对应的index,省去每次找到对应点的时间。
每个点访问一次,Time O(n);建立了HashMap,Space O(n).
Note: 1. Recursion终止条件是L>R 而不是L>=R,因为只有一个元素时L=R,此时应该返回元素而不是返回null。
2. 题目中没有给其他的 TreeNode constructor, 所以不能用,例如TreeNode() 就不可以。
3. 采用help时,argument 要传HashMap<Integer, Integer> 不能省略<Integer, Integer>. Or it would throw error like Objectcould not be assigned to int.
AC Java:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class Solution { public TreeNode buildTree(int[] preorder, int[] inorder) { if(preorder == null || preorder.length == 0 || inorder == null || inorder.length == 0 || preorder.length != inorder.length){ return null; } HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>(); for(int i = 0; i < inorder.length; i++){ hm.put(inorder[i],i); } return helper(preorder,0,preorder.length-1, inorder,0,inorder.length-1,hm); } private TreeNode helper(int[] preorder, int preL, int preR, int[] inorder, int inL, int inR, HashMap<Integer, Integer> hm){ //error!!! if(preL>preR || inL>inR){ return null; } TreeNode tn = new TreeNode(preorder[preL]); int index = hm.get(tn.val); tn.left = helper(preorder, preL+1, index-inL+preL,inorder,inL,index-1, hm); tn.right = helper(preorder, index-inL+preL+1, preR, inorder, index+1, inR, hm); return tn; }}
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