【LeetCode】 105. Construct Binary Tree from Preorder and Inorder Traversal 从前序与中序遍历序列构造二叉树(Medium)(JAVA)
题目地址: https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
题目描述:
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
题目大意
根据一棵树的前序遍历与中序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
解题方法
1、前序遍历的根节点在第一个;中序遍历的根节点在中间,找到根节点,就能分为左右子树
2、迭代更新完整的树
/**
* 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;
}
}
执行用时 : 3 ms, 在所有 Java 提交中击败了 81.03% 的用户
内存消耗 : 40.1 MB, 在所有 Java 提交中击败了 66.67% 的用户