Description

Given a binary tree, return the inorder traversal of its nodes’ values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?

Solutions

　　中序遍历二叉树。

1. 递归

　　用迭代的方式还是比较好解决：

# 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 inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        res = []
        
        def _inorderTraversal(node):
            if not node:
                return
            _inorderTraversal(node.left)
            res.append(node.val)
            _inorderTraversal(node.right)
        
        _inorderTraversal(root)
        return res
# Runtime: 20 ms, faster than 59.57% of Python online submissions for Binary Tree Inorder Traversal.
# Memory Usage: 11.9 MB, less than 8.96% of Python online submissions for Binary Tree Inorder Traversal.

　　不用内部函数，分开写的话：

# 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, res):
        if not root:
            return
        self._inorderTraversal(root.left, res)
        res.append(root.val)
        self._inorderTraversal(root.right, res)
        
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        self._inorderTraversal(root, res)
        return res
# Runtime: 24 ms, faster than 99.88% of Python3 online submissions for Binary Tree Inorder Traversal.
# Memory Usage: 13.9 MB, less than 6.56% of Python3 online submissions for Binary Tree Inorder Traversal.

2. DFS-迭代

　　之前整理过的不是很好理解的方式跟上述效果一样：

# 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 inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        stack = []
        node = root
        res = []
        while node or stack:
            if node:
                stack.append(node)
                node = node.left
            else:
                node = stack.pop()
                res.append(node.val)
                node = node.right
        return res
# Runtime: 20 ms, faster than 59.57% of Python online submissions for Binary Tree Inorder Traversal.
# Memory Usage: 11.9 MB, less than 8.96% of Python online submissions for Binary Tree Inorder Traversal.

　　找到一种比较好理解的方式：

# 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 inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        stack = []
        node = root
        res = []
        while node or stack:
            while node:
                stack.append(node)
                node = node.left
            node = stack.pop()
            res.append(node.val)
            node = node.right
        return res
# Runtime: 16 ms, faster than 81.93% of Python online submissions for Binary Tree Inorder Traversal.
# Memory Usage: 11.7 MB, less than 80.46% of Python online submissions for Binary Tree Inorder Traversal.

　　先一直遍历到左子树的最左边节点，其没有左孩子了，所以保存自己节点的值，然后遍历其右子树。注意中序遍历的时候

References

  1. 94. Binary Tree Inorder Traversal