105. 從前序與中序遍歷序列構造二叉樹-中等難度
阿新 • • 發佈:2020-07-12
問題描述
根據一棵樹的前序遍歷與中序遍歷構造二叉樹。
注意:
你可以假設樹中沒有重複的元素。
例如,給出
前序遍歷 preorder =[3,9,20,15,7]
中序遍歷 inorder = [9,3,15,20,7]
返回如下的二叉樹:
3
/ \
9 20
/ \
15 7
來源:力扣(LeetCode)
連結:https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal
解答
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * }*/ class Solution { public int find(int[] a,int key){ boolean flag = false; int index = 0; for(index=0;index<a.length;index++){ if(a[index]==key){ flag = true; break; } } if(flag)return index;return -1; } public TreeNode recursive(int[] pre, int[] in){ if(in.length==0)return null; if(in.length==1)return new TreeNode(in[0],null,null); TreeNode r = null; int position = find(in,pre[0]); TreeNode l = recursive(Arrays.copyOfRange(pre, 1, position+1),Arrays.copyOfRange(in, 0, position));if(position+1<in.length) r = recursive(Arrays.copyOfRange(pre, position+1, pre.length),Arrays.copyOfRange(in, position+1, in.length)); return new TreeNode(pre[0],l,r); } public TreeNode buildTree(int[] preorder, int[] inorder) { if(preorder.length==0)return null; return recursive(preorder,inorder); } }