做题博客链接
https://blog.csdn.net/qq_43349112/article/details/108542248
题目链接
https://leetcode-cn.com/problems/zhong-jian-er-cha-shu-lcof/
描述
输入某二叉树的前序遍历和中序遍历的结果，请重建该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复
的数字。
限制：
0 <= 节点个数 <= 5000
示例
例如，给出
前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：
3
/ \
9 20
/ \
15 7
初始代码模板
/**
* 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) {
}
}
代码
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
Map<Integer, Integer> map = new HashMap<>();
public TreeNode buildTree(int[] preorder, int[] inorder) {
for (int i = 0; i < inorder.length; i++) {
map.put(inorder[i], i);
}
return build(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1);
}
private TreeNode build(int[] preorder, int[] inorder, int ps, int pe, int is, int ie) {
//System.out.printf("%d %d %d %d\n", ps, pe, is, ie);
if (ps > pe || is > ie) {
return null;
}
TreeNode root = new TreeNode(preorder[ps]);
int index = map.get(preorder[ps]);
// 划分左右子树
root.left = build(preorder, inorder, ps + 1, ps + index - is, is, index - 1);
root.right = build(preorder, inorder, ps + index - is + 1, pe, index + 1, ie);
return root;
}
}