【LeetCode】 105. Construct Binary Tree from Preorder and Inorder Traversal Construct a binary tree (Medium) (JAVA) from preorder and inorder traversal sequence
Subject address: https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
Title description:
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
Topic
The binary tree is constructed according to the pre-order and middle-order traversal of a tree.
Note:
You can assume that there are no duplicate elements in the tree.
Problem solving method
1. The root node of the pre-order traversal is in the first one; the root node of the middle-order traversal is in the middle. If the root node is found, it can be divided into left and right subtrees.
2. Iteratively update the complete tree
/**
* 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> map = new HashMap<>();
for (int i = 0; i < inorder.length; i++) {
map.put(inorder[i], i);
}
return bH(preorder, inorder, map, 0, preorder.length - 1, 0, inorder.length - 1);
}
public TreeNode bH(int[] preorder, int[] inorder, Map<Integer, Integer> map, int pStart, int pEnd, int iStart, int iEnd) {
if (pStart > pEnd || iStart > iEnd) return null;
int mid = map.get(preorder[pStart]);
TreeNode root = new TreeNode(preorder[pStart]);
root.left = bH(preorder, inorder, map, pStart + 1, pStart + mid - iStart, iStart, mid - 1);
root.right = bH(preorder, inorder, map, pStart + mid - iStart + 1, pEnd, mid + 1, iEnd);
return root;
}
}
Execution time: 3 ms, defeated 81.03% of users
in all Java submissions Memory consumption: 40.1 MB, defeated 66.67% of users in all Java submissions