遍历二叉树

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⭐非递归实现

中序遍历

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class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        cur = root  #借助临时的cur，不要修改root
        res = []
        stack = []
        while cur or stack:
            if cur:  # 一直往左走，全部入栈
                stack.append(cur) 
                cur = cur.left
            else:  # 取出栈顶节点，处理，再处理该节点的右子树   
                cur = stack.pop()
                res.append(cur.val)  
                cur = cur.right
        return res

前序遍历

写法一

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class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        cur = root
        res = []
        stack = []
        while cur or stack:
            if cur:
                stack.append(cur)  # 每一个节点都放入栈中
                res.append(cur.val)
                cur = cur.left
            else:
                cur = stack.pop()  # 这里pop出来的是已经访问过的节点
                cur = cur.right  # 还需要指向右子树
        return res

❤写法二：

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class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        cur = root
        res = []
        stack = []
        while cur or stack:
            if cur:
                stack.append(cur.right)  # 栈中直接放入已访问节点的右子树
                res.append(cur.val)
                cur = cur.left
            else:
                cur = stack.pop()  # 取到的就是右子树根节点
        return res

😐写法三：

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class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        stack, output = [root], []
        
        while stack:
            root = stack.pop()
            if root is not None:
                output.append(root.val)
                if root.right is not None:  #右子树先入栈，左子树后入栈
                    stack.append(root.right)
                if root.left is not None:
                    stack.append(root.left)
        
        return output

后序遍历

https://leetcode-cn.com/problems/binary-tree-postorder-traversal/

对于当前节点，只有其右子树已经处理或没有右子树时，才能处理它。

👀这里要记录一个last_visited节点​​

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class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        cur = root
        res = []
        stack = []
        last_visited = None
        while cur or stack:
            if cur:
                stack.append(cur)
                cur = cur.left
            else:
                tmp = stack[-1]  # 定位根节点，这个地方不能pop(),因为可能有右子树
                if not tmp.right or tmp.right == last_visited:  
                    # 没有右子树或者已经访问过了，处理当前根节点
                    res.append(tmp.val)
                    last_visited = tmp
                    stack.pop()
                else:
                    cur = tmp.right  # 右子树未访问，处理右子树
        return res

❤后序遍历可以借前序遍历的一个变体实现，先根右左,再倒置。

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class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        cur = root
        res = []
        stack = []
        while cur or stack:
            if cur:
                stack.append(cur.left)  # 栈中直接放入已访问节点的左子树
                res.append(cur.val)
                cur = cur.right
            else:
                cur = stack.pop()  # 取到的就是左子树根节点
        return res[::-1]