面试题６－－利用前序和中序遍历重构二叉树－－递归方法

class TreeNode:
    def __init__(self,data,left,right):
        self.data=data
        self.left=left
        self.right=right
方法１
def construct_tree(pre_order,mid_order):
##保证结点数不为０，该树才能重构
    if (len(pre_order)==0) or (len(mid_order)==0):
        return None
##根节点即为前序遍历的首元素
    root_data=pre_order[0]# make sure the root_Data
 
##若结点个数为１，则不含左右子书
    if len(pre_order)==1 :
        return TreeNode(root_data,None,None)
##在中序遍历中查找根节点，确定其下标
    for i in range(len(mid_order)):
        if root_data==mid_order[i]:
            break
    root_mid_index=i
    #if root_mid_index >0:
＃＃分别重构其左右子书    
    left=construct_tree(pre_order[1:root_mid_index+1],mid_order[:root_mid_index])
    if root_mid_index<len(mid_order):
        right=construct_tree(pre_order[root_mid_index+1:],mid_order[root_mid_index+1:])
    return TreeNode(root_data,left,right)
    
if __name__=='__main__':
    pre_order = [1, 2, 4, 7, 3, 5, 6, 8]
    mid_order = [4, 7, 2, 1, 5, 3, 8, 6]
    root=construct_tree(pre_order,mid_order)

方法２
#def reConstructBinaryTree(pre, tin):
#    # write code here
#    if (len(pre) == 0) or (len(tin) == 0):
#        return None
#    
#    rootValue = pre[0]
#    root = TreeNode(rootValue,None,None)
#    if len(pre)==1:
#        return root
#    rootTinIndex = 0
#    for i in range(len(tin)):
#        if tin[i] == rootValue:
#            rootTinIndex = i
#    preStart = 1
#    preEnd = rootTinIndex+1
#    tinStart = 0
#    tinEnd = rootTinIndex
#    if rootTinIndex > 0:
＃＃用递归函数返回的结果赋值给某个变量　该变量之前必须定义过初值，如root.left/right==None
#        root.left = reConstructBinaryTree(pre[preStart:preEnd], tin[tinStart:tinEnd])
#    if rootTinIndex < len(pre):
#        root.right = reConstructBinaryTree(pre[preEnd:], tin[tinEnd+1:])
#    return root
#if __name__=='__main__':
#    pre = [1, 2, 4, 7, 3, 5, 6, 8]
#    tin = [4, 7, 2, 1, 5, 3, 8, 6]
#    root=reConstructBinaryTree(pre,tin)
    print root.data
    print root.left.data,root.right.data
    print root.left.left.data,root.right.left.data,root.right.right.data
    print root.left.left.right.data,root.right.right.left.data

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