LeetCode Construct Binary Tree from Preorder and Inorder Traversal
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LeetCode解题之Construct Binary Tree from Preorder and Inorder Traversal
原题
通过一棵二叉树的前序和中序排列来得出它的树形结构。
注意点:
- 无
例子:
输入: preorder = [3,9,20,15,14,7], inorder = [9,3,14,15,20,7]
输出:
3 / \ 9 20 / \ 15 7 / 14
解题思路
- 因为先序中的节点为根节点,所以取出先序的第一个节点
- 用先序的第一个节点可以将中序分成左右子树,然后取出先序的第二个节点将左右子树再次划分
- 当将中序全部划分为单个点时就结束
开始递归的时候用了列表作为参数,结果内存溢出,改用下标作参数,先序和中序序列作为全局变量。
AC源码
# Definition for a binary tree node.class TreeNode(object): def __init__(self, x): self.val = x self.left = None self.right = Noneclass Solution(object): def buildTree(self, preorder, inorder): """ :type preorder: List[int] :type inorder: List[int] :rtype: TreeNode """ self.preorder = preorder self.inorder = inorder return self._buildTree(0, len(preorder), 0, len(inorder)) def _buildTree(self, pre_start, pre_end, in_start, in_end): if pre_start >= pre_end: return None root = TreeNode(self.preorder[pre_start]) offset = self.inorder[in_start:in_end + 1].index(root.val) root.left = self._buildTree(pre_start + 1, pre_start + offset + 1, in_start, in_start + offset) root.right = self._buildTree(pre_start + offset + 1, pre_end, in_start + offset + 1, in_end) return rootif __name__ == "__main__": None
欢迎查看我的Github (https://github.com/gavinfish/LeetCode-Python) 来获得相关源码。
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