Binary Tree Traverasl with OO and Stack

1、一切都是對象，對象就是關系和操作；每個節點，都有兩種操作，print和visit。 2、放在stack里面可以使操作，可以是函數，遞歸也是面向對象的體現；

Python代碼實現如下：

class guide(object):def __init__(self, opt,node):self.opt = optself.node = nodeclass Solution2(object):def preorderTraversal(self, root):""":type root: TreeNode:rtype: List[int]"""if root is None:return []ans = []path = [guide(0,root),]while path:current = path.pop()if not current.node:continueif current.opt == 1:ans += [current.node.item]else:path += [guide(0,current.node.right)]path += [guide(0,current.node.left)]path += [guide(1,current.node)]return ansdef inorderTraversal(self, root):""":type root: TreeNode:rtype: List[int]"""if root is None:return []ans = []path = [guide(0,root),]while path:current = path.pop()if not current.node:continueif current.opt == 1:ans += [current.node.item]else:path += [guide(0,current.node.right)]path += [guide(1,current.node)]path += [guide(0,current.node.left)]return ansdef postorderTraversal(self, root):""":type root: TreeNode:rtype: List[int]"""if root is None:return []ans = []path = [guide(0,root),]while path:current = path.pop()if not current.node:continueif current.opt == 1:ans += [current.node.item]else:path += [guide(1,current.node)]path += [guide(0,current.node.right)]path += [guide(0,current.node.left)]return ans

Test case：

if __name__ =="__main__":tree =tree.Node()tree._add(0)tree._add(1)tree._add(2)tree._add(3)tree._add(4)tree._add(5)tree._add(6)tree._add(7)tree._add(8)tree._add(9)tree._add(10)ans = Solution3().preorderTraversal(tree)print ansans = Solution3().inorderTraversal(tree)print ansans = Solution3().postorderTraversal(tree)print ans

Output:

[0, 1, 3, 5, 7, 9, 10, 8, 6, 4, 2] [9, 7, 10, 5, 8, 3, 6, 1, 4, 0, 2] [9, 10, 7, 8, 5, 6, 3, 4, 1, 2, 0]

總結

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