/**
Given two integer arrays preorder and inorder where preorder is the preorder
traversal of a binary tree and inorder is the inorder traversal of the same tree,
construct and return the binary tree.
Example 1:
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]
Example 2:
Input: preorder = [-1], inorder = [-1]
Output: [-1]
Constraints:
1 <= preorder.length <= 3000
inorder.length == preorder.length
-3000 <= preorder[i], inorder[i] <= 3000
preorder and inorder consist of unique values.
Each value of inorder also appears in preorder.
preorder is guaranteed to be the preorder traversal of the tree.
inorder is guaranteed to be the inorder traversal of the tree.
Related Topics树 | 数组 | 哈希表 | 分治 | 二叉树
👍 1699, 👎 0
*/
//leetcode submit region begin(Prohibit modification and deletion)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
return buildTreeHelper(preorder, inorder, (long)Integer.MAX_VALUE + 1);
}
int pre = 0; int in = 0;
private TreeNode buildTreeHelper(int[] preorder, int[] inorder, long stop) {
//到达末尾返回 null
if (pre == preorder.length) {
return null;
}
if (inorder[in] == stop) {
in++;
return null;
}
int root_val = preorder[pre++];
TreeNode root = new TreeNode(root_val);
root.left = buildTreeHelper(preorder, inorder, root_val);
root.right = buildTreeHelper(preorder, inorder, stop);
return root;
}
}
