给定一棵二叉树，任务是计算二叉树中节点的数量，这些节点是从根到该节点的路径中价值最高的节点。
例子：
Input: Below is the given Tree:
3
/ \
2 5
/ \
4 6
Output: 4
Explanation:
Root node satisfies the required condition.
Node 5 is the highest valued node in the path (3 -> 5).
Node 6 is the highest valued node in the path (3 -> 5 -> 6).
Node 4 is the highest valued node in the path (3 -> 2 -> 4).
Therefore, there are a total of 4 such nodes in the given binary tree.
Input: Below is the given Tree:
3
/ \
1 2
/ \
4 6
Output: 3
方法：思想是对给定的二叉树进行中序遍历，并在每次递归调用后更新路径中迄今为止获得的最大值节点。请按照以下步骤操作：
- 对给定的二叉树执行中序遍历
- 每次递归调用后，更新从根节点到当前节点的路径中迄今为止遇到的最大值节点。
- 如果该节点的值超过目前路径中的最大值节点，则将计数增加1并更新到目前为止获得的最大值。
- 继续到当前节点的子树。
- 完全遍历二叉树后，打印得到的计数。
下面是上述方法的实现：
C++14
// C++14 program for the above approach
#include
using namespace std;
// Stores the ct of
// nodes which are maximum
// in the path from root
// to the current node
int ct = 0;
// Binary Tree Node
struct Node
{
int val;
Node *left, *right;
Node(int x)
{
val = x;
left = right = NULL;
}
};
// Function that performs Inorder
// Traversal on the Binary Tree
void find(Node *root, int mx)
{
// If root does not exist
if (root == NULL)
return;
// Check if the node
// satisfies the condition
if (root->val >= mx)
ct++;
// Update the maximum value
// and recursively traverse
// left and right subtree
find(root->left, max(mx, root->val));
find(root->right, max(mx, root->val));
}
// Function that counts the good
// nodes in the given Binary Tree
int NodesMaxInPath(Node* root)
{
// Perform inorder Traversal
find(root, INT_MIN);
// Return the final count
return ct;
}
// Driver code
int main()
{
/* A Binary Tree
3
/ \
2 5
/ \
4 6
*/
Node* root = new Node(3);
root->left = new Node(2);
root->right = new Node(5);
root->left->left = new Node(4);
root->right->right = new Node(7);
// Function call
int answer = NodesMaxInPath(root);
// Print the count of good nodes
cout << (answer);
return 0;
}
// This code is contributed by mohit kumar 29 Java
// Java program for the above approach
import java.util.*;
class GfG {
// Stores the count of
// nodes which are maximum
// in the path from root
// to the current node
static int count = 0;
// Binary Tree Node
static class Node {
int val;
Node left, right;
}
// Function that performs Inorder
// Traversal on the Binary Tree
static void find(Node root, int max)
{
// If root does not exist
if (root == null)
return;
// Check if the node
// satisfies the condition
if (root.val >= max)
count++;
// Update the maximum value
// and recursively traverse
// left and right subtree
find(root.left,
Math.max(max, root.val));
find(root.right,
Math.max(max, root.val));
}
// Function that counts the good
// nodes in the given Binary Tree
static int NodesMaxInPath(Node root)
{
// Perform inorder Traversal
find(root, Integer.MIN_VALUE);
// Return the final count
return count;
}
// Function that add the new node
// in the Binary Tree
static Node newNode(int data)
{
Node temp = new Node();
temp.val = data;
temp.left = null;
temp.right = null;
// Return the node
return temp;
}
// Driver Code
public static void main(String[] args)
{
/* A Binary Tree
3
/ \
2 5
/ \
4 6
*/
Node root = null;
root = newNode(3);
root.left = newNode(2);
root.right = newNode(5);
root.left.left = newNode(4);
root.right.right = newNode(7);
// Function Call
int answer = NodesMaxInPath(root);
// Print the count of good nodes
System.out.println(answer);
}
}Python3
# Python 3 program for the
# above approach
import sys
# Stores the ct of
# nodes which are maximum
# in the path from root
# to the current node
ct = 0
# Binary Tree Node
class newNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
# Function that performs Inorder
# Traversal on the Binary Tree
def find(root, mx):
global ct
# If root does not exist
if (root == None):
return
# Check if the node
# satisfies the condition
if (root.val >= mx):
ct += 1
# Update the maximum value
# and recursively traverse
# left and right subtree
find(root.left,
max(mx, root.val))
find(root.right,
max(mx, root.val))
# Function that counts
# the good nodes in the
# given Binary Tree
def NodesMaxInPath(root):
global ct
# Perform inorder
# Traversal
find(root,
-sys.maxsize-1)
# Return the final count
return ct
# Driver code
if __name__ == '__main__':
'''
/* A Binary Tree
3
/ /
2 5
/ /
4 6
*/
'''
root = newNode(3)
root.left = newNode(2)
root.right = newNode(5)
root.left.left = newNode(4)
root.right.right = newNode(7)
# Function call
answer = NodesMaxInPath(root)
# Print the count of good nodes
print(answer)
# This code is contributed by Surendra_GangwarC#
// C# program for
// the above approach
using System;
class GfG{
// Stores the count of
// nodes which are maximum
// in the path from root
// to the current node
static int count = 0;
// Binary Tree Node
public class Node
{
public int val;
public Node left,
right;
}
// Function that performs
// Inorder Traversal on
// the Binary Tree
static void find(Node root,
int max)
{
// If root does not exist
if (root == null)
return;
// Check if the node
// satisfies the condition
if (root.val >= max)
count++;
// Update the maximum value
// and recursively traverse
// left and right subtree
find(root.left,
Math.Max(max, root.val));
find(root.right,
Math.Max(max, root.val));
}
// Function that counts the good
// nodes in the given Binary Tree
static int NodesMaxInPath(Node root)
{
// Perform inorder Traversal
find(root, int.MinValue);
// Return the readonly count
return count;
}
// Function that add the new node
// in the Binary Tree
static Node newNode(int data)
{
Node temp = new Node();
temp.val = data;
temp.left = null;
temp.right = null;
// Return the node
return temp;
}
// Driver Code
public static void Main(String[] args)
{
/* A Binary Tree
3
/ \
2 5
/ \
4 6
*/
Node root = null;
root = newNode(3);
root.left = newNode(2);
root.right = newNode(5);
root.left.left = newNode(4);
root.right.right = newNode(7);
// Function Call
int answer = NodesMaxInPath(root);
// Print the count of good nodes
Console.WriteLine(answer);
}
}
// This code is contributed by Princi Singh输出：
4
时间复杂度： O(N)
辅助空间： O(1)
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