上次编辑到这里，代码来自缓存 点击恢复默认模板
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
}
};
python3 解法, 执行用时: 218 ms, 内存消耗: 86.4 MB, 提交时间: 2024-02-20 09:29:49
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
if len(preorder) == 0: return None
root = TreeNode(preorder[0]) # 根节点就是前序遍历的第一个节点
for i, node in enumerate(inorder): # 找到中序的根节点索引
if node == preorder[0]:
break
root.left = self.buildTree(preorder[1:i+1], inorder[:i])
root.right = self.buildTree(preorder[i+1:], inorder[i+1:])
return root
golang 解法, 执行用时: 8 ms, 内存消耗: 4 MB, 提交时间: 2022-11-13 11:37:04
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func buildTree(preorder []int, inorder []int) *TreeNode {
if len(preorder) == 0 {
return nil
}
root := &TreeNode{preorder[0], nil, nil}
i := 0
for ; i < len(inorder); i++ {
if inorder[i] == preorder[0] {
break
}
}
root.Left = buildTree(preorder[1:len(inorder[:i])+1], inorder[:i])
root.Right = buildTree(preorder[len(inorder[:i])+1:], inorder[i+1:])
return root
}
