算法题（七）--前后后序遍历，迭代版

class TreeNode():
    def __init__(self,val = 0,left = None,right = None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def inorderTraversal(self,root):
        def inner_find(node):
            while node:
                node_stack.append(node)
                node = node.left

        res = []; node_stack = []
        inner_find(root)   #找到所有左边的节点
        while node_stack:
            temp = node_stack.pop()
            res.append(temp.val)
            inner_find(temp.right)  #找到所有右边的节点
        return res

    def forwardTraversal(self,root):
        res = [];
        node_stack = []
        node_stack.append(root)
        while node_stack:
            temp = node_stack.pop()
            res.append(temp.val)
            if temp.right:
                node_stack.append(temp.right)
            if temp.left:
                node_stack.append(temp.left)
        return res

    def behindTraversal(self,root):
        def inner_find(node):
            while node:
                node_stack.append(node)
                node = node.left
        res = []; node_stack = [];pre = None
        inner_find(root)   #找到所有左边的节点
        while node_stack:
            flage = node_stack[-1]
            if flage.right == None or flage.right ==pre:
                temp = node_stack.pop()
                res.append(temp.val)
                pre = temp
            else:
                inner_find(flage.right)
        return res

'''
前序遍历最简单
中序遍历和后序遍历有相同点，都是先找最左边的节点，不同的是中序遍历是找完最左边节点之后就直接取值，再去找右边的节点，再找右边节点；
而后续遍历需要增加一些苛刻的条件才可以取值，这个条件不满足就不能取值，而是因为继续找右边的节点。
'''

if __name__ == '__main__': #构建一棵树 #1、先创建点 A,B,C,D,E,F,G,H,I = [TreeNode(x) for x in 'ABCDEFGHI'] #2、再创建树的依赖关系 A.left,A.right = B,C B.right = D C.left,C.right = E,F E.left = G F.left,F.right = H,I #3、对树进行中序遍历，即 B,D,A,G,E,C,H,F,I print(Solution().inorderTraversal(A)) #4、对数进行前序遍历，即A,B,D,C,E,G,F,H,I print(Solution().forwardTraversal(A)) #5、对数进行后序遍历，即D,B,G,E,H,I,F,C,A print(Solution().behindTraversal(A))