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Difficulty
Medium.
Problem
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
AC
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# DFS
class Solution():
def buildTree(self, inorder, postorder):
if not len(postorder):
return None
root = TreeNode(postorder[-1])
idx = inorder.index(root.val)
root.left = self.buildTree(inorder[ : idx], postorder[ : idx])
root.right = self.buildTree(inorder[idx+1 : ], postorder[idx : -1])
return root