边工作边刷题：70天一遍leetcode: day 16

Flatten Binary Tree to Linked List

同理，这题递归的解法trivial，试着用iterative解了下，后续遍历，然后找左子树的右孩子。然后…就不需要然后了

# 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 flatten(self, root):
        """
        :type root: TreeNode
        :rtype: void Do not return anything, modify root in-place instead.
        """
        if not root: return 
        stk = [root]
        pre = None
        while stk:
            cur = stk[-1]
            if not pre or pre.left==cur or pre.right==cur:
                if cur.left:
                    stk.append(cur.left)
                elif cur.right:
                    stk.append(cur.right)
            elif cur.left==pre:
                if cur.right:
                    stk.append(cur.right)
            else:
                stk.pop()
                if cur.left:
                    curl = cur.left
                    while curl.right:
                        curl = curl.right
                    curl.right = cur.right
                    cur.right = cur.left
                    cur.left = None
            pre = cur