在二叉树中查找从根到给定节点的距离
给定二叉树的根和其中的键 x，求给定键到根的距离。距离是指两个节点之间的边数。
例子：
Input : x = 45,
Root of below tree
5
/ \
10 15
/ \ / \
20 25 30 35
\
45
Output : Distance = 3
There are three edges on path
from root to 45.
For more understanding of question,
in above tree distance of 35 is two
and distance of 10 is 1.方法：这个想法是从根遍历树。检查 x 是否存在于根或左子树或右子树中。我们将距离初始化为 -1，并对所有三种情况的距离加 1。
C++
// C++ program to find distance of a given
// node from root.
#include
using namespace std;
// A Binary Tree Node
struct Node
{
int data;
Node *left, *right;
};
// A utility function to create a new Binary
// Tree Node
Node *newNode(int item)
{
Node *temp = new Node;
temp->data = item;
temp->left = temp->right = NULL;
return temp;
}
// Returns -1 if x doesn't exist in tree. Else
// returns distance of x from root
int findDistance(Node *root, int x)
{
// Base case
if (root == NULL)
return -1;
// Initialize distance
int dist = -1;
// Check if x is present at root or in left
// subtree or right subtree.
if ((root->data == x) ||
(dist = findDistance(root->left, x)) >= 0 ||
(dist = findDistance(root->right, x)) >= 0)
return dist + 1;
return dist;
}
// Driver Program to test above functions
int main()
{
Node *root = newNode(5);
root->left = newNode(10);
root->right = newNode(15);
root->left->left = newNode(20);
root->left->right = newNode(25);
root->left->right->right = newNode(45);
root->right->left = newNode(30);
root->right->right = newNode(35);
cout << findDistance(root, 45);
return 0;
} Java
// Java program to find distance of a given
// node from root.
import java.util.*;
class GfG {
// A Binary Tree Node
static class Node
{
int data;
Node left, right;
}
// A utility function to create a new Binary
// Tree Node
static Node newNode(int item)
{
Node temp = new Node();
temp.data = item;
temp.left = null;
temp.right = null;
return temp;
}
// Returns -1 if x doesn't exist in tree. Else
// returns distance of x from root
static int findDistance(Node root, int x)
{
// Base case
if (root == null)
return -1;
// Initialize distance
int dist = -1;
// Check if x is present at root or in left
// subtree or right subtree.
if ((root.data == x) ||
(dist = findDistance(root.left, x)) >= 0 ||
(dist = findDistance(root.right, x)) >= 0)
return dist + 1;
return dist;
}
// Driver Program to test above functions
public static void main(String[] args)
{
Node root = newNode(5);
root.left = newNode(10);
root.right = newNode(15);
root.left.left = newNode(20);
root.left.right = newNode(25);
root.left.right.right = newNode(45);
root.right.left = newNode(30);
root.right.right = newNode(35);
System.out.println(findDistance(root, 45));
}
}Python3
# Python3 program to find distance of
# a given node from root.
# A class to create a new Binary
# Tree Node
class newNode:
def __init__(self, item):
self.data = item
self.left = self.right = None
# Returns -1 if x doesn't exist in tree.
# Else returns distance of x from root
def findDistance(root, x):
# Base case
if (root == None):
return -1
# Initialize distance
dist = -1
# Check if x is present at root or
# in left subtree or right subtree.
if (root.data == x):
return dist + 1
else:
dist = findDistance(root.left, x)
if dist >= 0:
return dist + 1
else:
dist = findDistance(root.right, x)
if dist >= 0:
return dist + 1
return dist
# Driver Code
if __name__ == '__main__':
root = newNode(5)
root.left = newNode(10)
root.right = newNode(15)
root.left.left = newNode(20)
root.left.right = newNode(25)
root.left.right.right = newNode(45)
root.right.left = newNode(30)
root.right.right = newNode(35)
print(findDistance(root, 45))
# This code is contributed by PranchalKC#
// C# program to find distance of a given
// node from root.
using System;
class GfG
{
// A Binary Tree Node
class Node
{
public int data;
public Node left, right;
}
// A utility function to create
// a new Binary Tree Node
static Node newNode(int item)
{
Node temp = new Node();
temp.data = item;
temp.left = null;
temp.right = null;
return temp;
}
// Returns -1 if x doesn't exist in tree. Else
// returns distance of x from root
static int findDistance(Node root, int x)
{
// Base case
if (root == null)
return -1;
// Initialize distance
int dist = -1;
// Check if x is present at root or in left
// subtree or right subtree.
if ((root.data == x) ||
(dist = findDistance(root.left, x)) >= 0 ||
(dist = findDistance(root.right, x)) >= 0)
return dist + 1;
return dist;
}
// Driver code
public static void Main(String[] args)
{
Node root = newNode(5);
root.left = newNode(10);
root.right = newNode(15);
root.left.left = newNode(20);
root.left.right = newNode(25);
root.left.right.right = newNode(45);
root.right.left = newNode(30);
root.right.right = newNode(35);
Console.WriteLine(findDistance(root, 45));
}
}
// This code is contributed by 29AjayKumarJavascript
输出：
3时间复杂度： O(N)
辅助空间： O(1)