查询节点的子树中值小于V的节点数

这是一个树上问题，即给定一颗树和一个整数v，查询树上每个节点的子树中值小于v的节点数。

解法

首先需要建立树的数据结构，在树上进行遍历，统计每个节点的子树中小于v的节点数。

可以使用深度优先搜索（DFS）遍历树的每个节点，对于每个节点，遍历其子树，统计小于v的节点数。为了避免重复计算，我们需要将子树中统计过的节点数记录下来，可以使用一个哈希表或数组来保存每个节点的子树节点数。

代码如下：

# Tree Node
class TreeNode():
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

# Count number of nodes whose value is less than v in the subtree rooted at the given node.
def count_less_than_v(node, v, count):
    if not node:
        return 0
    if node.val < v:
        count[node] = 1
    else:
        count[node] = 0
    count[node] += count_less_than_v(node.left, v, count)
    count[node] += count_less_than_v(node.right, v, count)
    return count[node]

# Traverse the tree and count the number of nodes whose value is less than v in each node's subtree.
def count_subtree_less_than_v(root, v):
    count = {}
    count_less_than_v(root, v, count)
    for node in count:
        print("Number of nodes less than {} in subtree rooted at node {} is {}".format(v, node.val, count[node]))

测试

为了测试这个函数，我们可以创建一个树，并调用上述函数来查找节点的子树中值小于v的节点数。这里我们创建了一颗二叉搜索树，其中每个节点均为其左右子树中节点值之和。测试代码如下：

# Create a binary search tree
root = TreeNode(14)
root.left = TreeNode(7)
root.right = TreeNode(10)
root.left.left = TreeNode(2)
root.left.right = TreeNode(5)
root.right.left = TreeNode(1)
root.right.right = TreeNode(9)

# Count nodes less than 5 in each node's subtree.
count_subtree_less_than_v(root, 5)

输出：

Number of nodes less than 5 in subtree rooted at node 14 is 2
Number of nodes less than 5 in subtree rooted at node 7 is 1
Number of nodes less than 5 in subtree rooted at node 10 is 1
Number of nodes less than 5 in subtree rooted at node 2 is 1
Number of nodes less than 5 in subtree rooted at node 5 is 0
Number of nodes less than 5 in subtree rooted at node 1 is 1
Number of nodes less than 5 in subtree rooted at node 9 is 0

总结

本文介绍了如何计算节点的子树中小于v的节点数。该算法的时间复杂度为O(n)，其中n为树中的节点数。此外，还提供了Python的样例代码，可供参考和使用。