【题目】
给定二叉搜索树的根结点 root,返回值位于范围 [low, high] 之间的所有结点的值的和。
示例 1:
输入:root = [10,5,15,3,7,null,18], low = 7, high = 15
输出:32
示例 2:
输入:root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10
输出:23
提示:
树中节点数目在范围 [1, 2 * 104] 内
1 <= Node.val <= 105
1 <= low <= high <= 105
所有 Node.val 互不相同
【代码】
【方法1:深度优先】
class Solution:
def __init__(self):
self.s=0
def rangeSumBST(self, root: TreeNode, low: int, high: int) -> int:
if not root:
return 0
if root.val>=low and root.val<=high:
self.s+=root.val
self.rangeSumBST(root.left,low,high)
self.rangeSumBST(root.right,low,high)
return self.s
【方法2:广度优先遍历】
class Solution:
def rangeSumBST(self, root: TreeNode, low: int, high: int) -> int:
if not root:
return 0
s=0
que=[root]
while que:
node=que.pop(0)
if node.val>=low and node.val<=high:
s+=node.val
if node.left:
que.append(node.left)
if node.right:
que.append(node.right)
return s
【方法3:广度+剪枝】
class Solution:
def rangeSumBST(self, root: TreeNode, low: int, high: int) -> int:
if not root:
return 0
s=0
que=[root]
while que:
node=que.pop(0)
if node.val>=low and node.val<=high:
s+=node.val
if node.left:
que.append(node.left)
if node.right:
que.append(node.right)
elif node.val<low and node.right:
que.append(node.right)
elif node.val>high and node.left:
que.append(node.left)
return s
【方法4:深度+剪枝】
class Solution:
def __init__(self):
self.s=0
def rangeSumBST(self, root: TreeNode, low: int, high: int) -> int:
if not root:
return 0
if root.val>=low and root.val<=high:
self.s+=root.val
self.rangeSumBST(root.left,low,high)
self.rangeSumBST(root.right,low,high)
elif root.val<low and root.right:
self.rangeSumBST(root.right,low,high)
elif root.val>high and root.left:
self.rangeSumBST(root.left,low,high)
return self.s
【写法5】
class Solution:
def rangeSumBST(self, root: TreeNode, low: int, high: int) -> int:
if not root:
return 0
if root.val > high:
return self.rangeSumBST(root.left, low, high)
if root.val < low:
return self.rangeSumBST(root.right, low, high)
return root.val + self.rangeSumBST(root.left, low, high) + self.rangeSumBST(root.right, low, high)