原題

給出一棵二叉樹,返回其中序遍歷

給出二叉樹 {1,#,2,3}

   1
    \
     2
    /
   3

返回 [1,3,2]

解題思路

• 方法一：遞歸，定義helper函數
• 方法二：Divide & Conquer
• 方法二：非遞歸
• (操作一)每一次從當前節點開始(第一次是root節點)遍歷至最左節點，一次入棧。
• (操作二)pop出棧一個node，node.val加入到結果，然后看該node有沒有右兒子
• 有的話重復操作一
• 沒有的話，繼續pop出棧一個node，重復操作二
• 非遞歸解法可以體會一下另外一個題[Inorder Successor in Binary Search Tree]

完整代碼

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

# 方法一
class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        res = []
        self.helper(root, res)
        return res
        
    def helper(self, root, result):
        if root != None:
            self.helper(root.left, result)
            result.append(root.val)
            self.helper(root.right, result)

# 方法二
class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        result = []
        if root == None:
            return result
        result.append(root.val)
        
        # divide
        left = self.inorderTraversal(root.left)
        right = self.inorderTraversal(root.right)
        
        # conquer
        return left + result + right

# 方法三
class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        stack = []
        result = []
        current = root
        while current != None or len(stack) != 0:
            while current != None:
                stack.append(current)
                current = current.left
            current = stack[-1];
            stack.pop();
            result.append(current.val);
            current = current.right;
        return result