LeetCode Construct Binary Tree from Inorder and Postorder Traversal

LeetCode解题之Construct Binary Tree from Inorder and Postorder Traversal

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原题

通过一棵二叉树的中序和后序排列来得出它的树形结构。

注意点：

• 无

例子:

输入: inorder = [9,3,14,15,20,7], postorder = [9,14,15,7,20,3]

输出:

    3   / \  9  20    /  \   15   7  / 14

解题思路

1. 因为后序中的节点为根节点,所以取出后序的最后一个节点
2. 用后序的最后一个节点可以将中序分成左右子树，然后取出后序的倒数第二个节点将左右子树再次划分
3. 当将中序全部划分为单个点时就结束

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, inorder, postorder):        """        :type inorder: List[int]        :type postorder: List[int]        :rtype: TreeNode        """        self.postorder = postorder        self.inorder = inorder        return self._buildTree(0, len(inorder))    def _buildTree(self, start, end):        if start < end:            root = TreeNode(self.postorder.pop())            index = self.inorder.index(root.val)            root.right = self._buildTree(index + 1, end)            root.left = self._buildTree(start, index)            return rootif __name__ == "__main__":    None

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