leetcode 每日一题 94. 二叉树的中序遍历

递归

思路:

先遍历左子树，再访问根节点，再遍历右子树。

代码：

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

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        def helper(root,res):
            if root:
                if root.left:
                    helper(root.left,res)
                res.append(root.val)
                if root.right:
                    helper(root.right,res)
        res = []
        helper(root,res)
        return res

迭代

思路:

创建一个栈，从根节点遍历左子树，将对应根节点和左子树压入栈中，直到左子树为空。此时取出栈顶节点，将对应节点的值记录，继续对取出节点的右子树进行压栈和取出记录操作，取出的栈顶元素如果右子树为空，则记录后继续取出栈顶节点。

代码：

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

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        stack = []
        cur = root
        while cur or stack:
            while cur:
                stack.append(cur)
                cur = cur.left
            cur = stack.pop()
            res.append(cur.val)
            cur = cur.right 
        return res