LeetCode 105 Construct Binary Tree from Preorder and Inorder Traversal
此题难度中等
LeetCode题解专栏:LeetCode题解
我做的所有的LeetCode的题目都放在这个专栏里,大部分题目Java和Python的解法都有。
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
python递归解法:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
if len(preorder)==0:
return None
root=TreeNode(preorder[0])
middle=inorder.index(preorder[0])
root.left=self.buildTree(preorder[1:middle+1],inorder[:middle])
root.right=self.buildTree(preorder[middle+1:],inorder[middle+1:])
return root
java递归解法:
/**
* 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) {
if (preorder.length == 0) {
return null;
}
TreeNode root = new TreeNode(preorder[0]);
int middle = 0;
for (int i = 0; i < preorder.length; i++) {
if (inorder[i] == root.val) {
middle = i;
break;
}
}
root.left = buildTree(Arrays.copyOfRange(preorder, 1, middle + 1),
Arrays.copyOfRange(inorder, 0, middle));
root.right = buildTree(Arrays.copyOfRange(preorder, middle + 1, preorder.length),
Arrays.copyOfRange(inorder, middle + 1, preorder.length));
return root;
}
}