二叉树中序遍历

1、题目

图1

2、题解

递归

不断先询问左子树，然后将根节点值加入答案，再询问右子树，直到节点为空结束。

# Definition for a binary tree node.
# 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: Optional[TreeNode]) -> List[int]:
        res = []
        self.inorder(root,res)
        return res

    def inorder(self,root,res):
        if not root:
            return
        self.inorder(root.left,res)
        res.append(root.val)
        self.inorder(root.right,res)

迭代（栈）

定义一个栈，将树的左节点依次入栈，当左节点为空时，弹出栈顶元素，将该元素值加入列表，遍历右节点。

# Definition for a binary tree node.
# 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: Optional[TreeNode]) -> List[int]:
        res = []
        stack = []
        while root or stack:
            if root:
                stack.append(root)
                root =root.left
            else:
                root = stack.pop()
                res.append(root.val)
                root = root.right

        return res