【LeetCode】623. Add a row to the binary tree

题目

623. 在二叉树中增加一行

给定一个二叉树的根 root 和两个整数 val 和 depth ,在给定的深度 depth 处添加一个值为 val 的节点行.

注意,根节点 root 位于深度 1 .

加法规则如下:

• 给定整数 depth,对于深度为 depth - 1 的每个非空树节点 cur ,创建两个值为 val 的树节点作为 cur 的左子树根和右子树根.
• cur 原来的左子树应该是新的左子树根的左子树.
• cur 原来的右子树应该是新的右子树根的右子树.
• 如果 depth == 1 意味着 depth - 1 根本没有深度,那么创建一个树节点,值 val 作为整个原始树的新根,而原始树就是新根的左子树.

示例 1:

输入: root = [4,2,6,3,1,5], val = 1, depth = 2
输出: [4,1,1,2,null,null,6,3,1,5]

示例 2:

输入: root = [4,2,null,3,1], val = 1, depth = 3
输出:  [4,2,null,1,1,3,null,null,1]

提示:

• 节点数在 [1, 104] 范围内
• 树的深度在 [1, 104]范围内
• -100 <= Node.val <= 100
• -105 <= val <= 105
• 1 <= depth <= the depth of tree + 1

题解1

思路

• 若为根节点,The new node is directly used as the root node,The root node is returned as the left subtree of the new node
• if not the root node,则遍历到$depth-1$层,Then insert the new node

代码

# 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 addOneRow(self, root: Optional[TreeNode], val: int, depth: int) -> Optional[TreeNode]:
        if depth == 1:
            node = TreeNode(val)
            node.left = root
            root = node
            return root
        queue = collections.deque([root])
        while queue and depth > 2:
            depth -= 1
            for _ in range(len(queue)):
                cur = queue.popleft()
                if cur.left: queue.append(cur.left)
                if cur.right: queue.append(cur.right)
        for _ in range(len(queue)):
            cur = queue.popleft()
            if cur.left:
                node = TreeNode(val)
                node.left = cur.left
                cur.left = node
            else:
                cur.left = TreeNode(val)
            if cur.right:
                node = TreeNode(val)
                node.right = cur.right
                cur.right = node
            else:
                cur.right = TreeNode(val)
        return root
                

复杂度

• 时间复杂度：$O(n)$
• 空间复杂度：$O(n)$

题解2

思路

• 当$depth\in 1,2$时,直接插入新节点,Otherwise recurse its left and right subtrees,直到符合条件

代码

# 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 addOneRow(self, root: Optional[TreeNode], val: int, depth: int) -> Optional[TreeNode]:
        if not root: return None
        if depth == 1:
            return TreeNode(val,root,None)
        if depth == 2:
            root.left = TreeNode(val,root.left, None)
            root.right = TreeNode(val, None, root.right)
        else:
            root.left = self.addOneRow(root.left, val, depth-1)
            root.right = self.addOneRow(root.right, val, depth-1)
        return root
                

复杂度

• 时间复杂度：$O(n)$
• 空间复杂度：$O(n)$