标签:cat amp ret map ons class turn integer note
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
/** * 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) { Map<Integer, Integer> mymap = new HashMap<>(); for (int i = 0; i < inorder.length; i++) { mymap.put(inorder[i], i); } return helper(0, inorder.length - 1, 0, preorder.length - 1, preorder, mymap); } private TreeNode helper(int inLeft, int inRight, int preLeft, int preRight, int[] preorder, Map<Integer, Integer> mymap) { if (inLeft > inRight) { return null; } TreeNode cur = new TreeNode(preorder[preLeft]); int leftSize = mymap.get(preorder[preLeft]) - inLeft; // preRight for left just add up leftSize cur.left = helper(inLeft, inLeft + leftSize - 1, preLeft + 1, preLeft + leftSize, preorder, mymap); cur.right = helper(inLeft + leftSize + 1, inRight, preLeft + leftSize + 1, preRight, preorder, mymap); return cur; } }
[LC] 105. Construct Binary Tree from Preorder and Inorder Traversal
标签:cat amp ret map ons class turn integer note
原文地址:https://www.cnblogs.com/xuanlu/p/12170809.html