如何从已知的遍历次序构建二叉树

思路

  1. 中序 + 前序
    由前序确定根结点,再由中序通过根节点分为左子树和右子树,之后分别对子树递归
  2. 中序 + 后序
    由后序确定根结点,再由中序通过根节点分为左子树和右子树,之后分别对子树递归

以下为Leetcode题解的Python代码

106.Construct Binary Tree from Inorder and Postorder Traversal

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def buildTree(self, inorder, postorder):
        """
        :type inorder: List[int]
        :type postorder: List[int]
        :rtype: TreeNode
        """
        if len(inorder) == 0 or len(postorder) == 0:
            return None
        val = postorder[-1]
        node = TreeNode(val)
        idx = inorder.index(val)

        left = self.buildTree(inorder[:idx], postorder[:idx])
        right = self.buildTree(inorder[idx+1:], postorder[-len(inorder)+idx:-1])
        node.left = left
        node.right = right
        return node

105. Construct Binary Tree from Preorder and Inorder Traversal

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def buildTree(self, preorder, inorder):
        """
        :type preorder: List[int]
        :type inorder: List[int]
        :rtype: TreeNode
        """
        if len(inorder) == 0 or len(preorder) == 0:
            return None
        val = preorder[0]
        node = TreeNode(val)
        idx = inorder.index(val)
        left = self.buildTree(preorder[1:idx+1], inorder[:idx])
        right = self.buildTree(preorder[idx+1:], inorder[idx+1:])
        node.left = left
        node.right = right
        return node

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