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

 线索二叉树

思路:

创建两个指针cur和pre，分别代表当前节点，和当前节点的前驱，由中序遍历的性质得：

①如果cur没有左子节点，将cur的val添加到输出，继续遍历右子树即cur = cur.right

②如果cur有左子节点，则找到左子树的最右侧节点的右子节点为pre，记录cur的左子节点temp，令cur的左子节点为空，pre的右子节点为cur，cur变为temp

只要cur不为空就继续①②操作

例如:

 

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 = []
        cur,pre = root,None
        while cur:
            if not cur.left:
                res.append(cur.val)
                cur = cur.right
            else:
                pre = cur.left
                while pre.right:
                    pre = pre.right
                pre.right = cur
                temp = cur
                cur = cur.left
                temp.left = None
        return res

 线索二叉树:(不改变树的结构)

思路:

在上面二叉树线索二叉树的基础上，可以在当前节点cur和当前节点的前驱pre之间设置一个临时的链接：pre.right = cur。在当前cur节点找它的前驱pre时：

①如果没有链接，设置链接并走向左子树，pre.right = cur , cur = cur.left

②如果有链接，断开链接并走向右子树，pre.right = None , cur = cur.right

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 = []
        cur,pre = root,None
        while cur:
            if not cur.left:
                res.append(cur.val)
                cur = cur.right
            else:
                pre = cur.left
                while pre.right and pre.right!=cur:
                    pre = pre.right
                if pre.right is None:
                    pre.right = cur
                    cur = cur.left
                else:
                    res.append(cur.val)
                    pre.right = None
                    cur = cur.right  
        return res