题目来源:

  https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/


题意分析:

  根据二叉树的中序和前序遍历结果,反推这个树,假设只有一个这样的树。


题目思路:

  前序遍历的第一个树将中序遍历分成两半,左半部分是左子树,右半部分是右子树,递归重构这棵树。


代码(python):

  

# 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
"""
def dfs(pbegin,pend,ibegin,iend):
if pbegin >= pend:
return None
if pbegin + 1 == pend:
return TreeNode(preorder[pbegin])
i = inorder.index(preorder[pbegin])
i -= ibegin
ans = TreeNode(preorder[pbegin])
ans.left = dfs(pbegin + 1,pbegin + i + 1,ibegin,ibegin+i)
ans.right = dfs(pbegin + i + 1,pend,ibegin + 1 + i,iend)
return ans
return dfs(0,len(preorder),0,len(inorder))

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