给定一个由N 个节点组成的二叉树，其值在[1, N]范围内，任务是找到从根节点到树的每个节点的距离。
例子：
Input:
Output: 0 1 1 2 2 2
Explanation:
The distance from the root to node 1 is 0.
The distance from the root to node 2 is 1.
The distance from the root to node 3 is 1.
The distance from the root to node 4 is 2.
The distance from the root to node 5 is 2.
The distance from the root to node 6 is 2.
Input:
Output: 3 3 2 1 0 1 2
方法：该问题可以使用 BFS 技术解决。这个想法是利用这样一个事实，即从根到节点的距离等于二叉树中该节点的级别。请按照以下步骤解决问题：
- 初始化一个队列，比如Q ，以存储树的每一层的节点。
- 初始化一个数组，比如dist[] ，其中dist[i]存储从根节点到树的第i个节点的距离。
- 使用 BFS 遍历树。对于遇到的每个第i个节点，将dist[i]更新为二叉树中该节点的级别。
- 最后，打印dist[]数组。
下面是上述方法的实现：
C++
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2 3
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4 5 6Java
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4 6
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1 2Python3
// C++ program to implement
// the above approach
#include
using namespace std;
struct Node {
// Stores data value
// of the node
int data;
// Stores left subtree
// of a node
Node* left;
// Stores right subtree
// of a node
Node* right;
Node(int x)
{
data = x;
left = right = NULL;
}
};
void findDistance(Node* root, int N)
{
// Store nodes at each level
// of the binary tree
queue Q;
// Insert root into Q
Q.push(root);
// Stores level of a node
int level = 0;
// dist[i]: Stores the distance
// from root node to node i
int dist[N + 1];
// Traverse tree using BFS
while (!Q.empty()) {
// Stores count of nodes
// at current level
int M = Q.size();
// Traverse the nodes at
// current level
for (int i = 0; i < M; i++) {
// Stores front element
// of the queue
root = Q.front();
// Pop front element
// of the queue
Q.pop();
// Stores the distance from
// root node to current node
dist[root->data] = level;
if (root->left) {
// Push left subtree
Q.push(root->left);
}
if (root->right) {
// Push right subtree
Q.push(root->right);
}
}
// Update level
level += 1;
}
for (int i = 1; i <= N; i++) {
cout << dist[i] << " ";
}
}
// Driver Code
int main()
{
int N = 7;
Node* root = new Node(5);
root->left = new Node(4);
root->right = new Node(6);
root->left->left = new Node(3);
root->left->right = new Node(7);
root->left->left->left = new Node(1);
root->left->left->right = new Node(2);
findDistance(root, N);
} C#
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
static class Node
{
// Stores data value
// of the node
int data;
// Stores left subtree
// of a node
Node left;
// Stores right subtree
// of a node
Node right;
Node(int x)
{
data = x;
left = right = null;
}
};
static void findDistance(Node root, int N)
{
// Store nodes at each level
// of the binary tree
Queue Q = new LinkedList<>();
// Insert root into Q
Q.add(root);
// Stores level of a node
int level = 0;
// dist[i]: Stores the distance
// from root node to node i
int []dist = new int[N + 1];
// Traverse tree using BFS
while (!Q.isEmpty())
{
// Stores count of nodes
// at current level
int M = Q.size();
// Traverse the nodes at
// current level
for (int i = 0; i < M; i++)
{
// Stores front element
// of the queue
root = Q.peek();
// Pop front element
// of the queue
Q.remove();
// Stores the distance from
// root node to current node
dist[root.data] = level;
if (root.left != null)
{
// Push left subtree
Q.add(root.left);
}
if (root.right != null)
{
// Push right subtree
Q.add(root.right);
}
}
// Update level
level += 1;
}
for (int i = 1; i <= N; i++)
{
System.out.print(dist[i] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int N = 7;
Node root = new Node(5);
root.left = new Node(4);
root.right = new Node(6);
root.left.left = new Node(3);
root.left.right = new Node(7);
root.left.left.left = new Node(1);
root.left.left.right = new Node(2);
findDistance(root, N);
}
}
// This code is contributed by shikhasingrajput 输出：
# Python3 program to implement
# the above approach
class Node:
def __init__(self, data):
# Stores data value
# of the node
self.data = data
# Stores left subtree
# of a node
self.left = None
# Stores right subtree
# of a node
self.right = None
def findDistance(root, N):
# Store nodes at each level
# of the binary tree
Q = []
# Insert root into Q
Q.append(root)
# Stores level of a node
level = 0
# dist[i]: Stores the distance
# from root node to node i
dist = [0 for i in range(N + 1)]
# Traverse tree using BFS
while Q:
# Stores count of nodes
# at current level
M = len(Q)
# Traverse the nodes at
# current level
for i in range(0, M):
# Stores front element
# of the queue
root = Q[0]
# Pop front element
# of the queue
Q.pop(0)
# Stores the distance from
# root node to current node
dist[root.data] = level
if root.left:
# Push left subtree
Q.append(root.left)
if root.right:
# Push right subtree
Q.append(root.right)
# Update level
level += 1
for i in range(1, N + 1):
print(dist[i], end = " ")
# Driver code
if __name__ == '__main__':
N = 7
root = Node(5)
root.left = Node(4)
root.right = Node(6)
root.left.left = Node(3)
root.left.right = Node(7)
root.left.left.left = Node(1)
root.left.left.right = Node(2)
findDistance(root, N)
# This code is contributed by MuskanKalra1时间复杂度： O(N)
辅助空间： O(N)
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