二叉树对角遍历中的第K个节点
给定一个二叉树和一个值K 。任务是打印二叉树对角遍历中的第 k 个节点。如果不存在这样的节点，则打印 -1。
例子：
Input :
8
/ \
3 10
/ / \
1 6 14
/ \ /
4 7 13
k = 5
Output : 6
Diagonal Traversal of the above tree is:
8 10 14
3 6 7 13
1 4
Input :
1
/ \
2 3
/ \
4 5
k = 7
Output : -1方法：思想是对二叉树进行对角遍历，直到在对角遍历中访问到K个节点。在遍历每个访问的节点时，递减变量K的值，并在 K 的值变为零时返回当前节点。如果对角线遍历不包含至少 K 个节点，则返回 -1。
下面是上述方法的实现：
C++
// C++ program to print kth node
// in the diagonal traversal of a binary tree
#include
using namespace std;
// A binary tree node has data, pointer to left
// child and a pointer to right child
struct Node {
int data;
Node *left, *right;
};
// Helper function that allocates a new node
Node* newNode(int data)
{
Node* node = new Node();
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Iterative function to print kth node
// in diagonal traversal of binary tree
int diagonalPrint(Node* root, int k)
{
// Base cases
if (root == NULL || k == 0)
return -1;
int ans = -1;
queue q;
// Push root node
q.push(root);
// Push delimiter NULL
q.push(NULL);
while (!q.empty()) {
Node* temp = q.front();
q.pop();
if (temp == NULL) {
if (q.empty()) {
// If kth node exists then return
// the answer
if (k == 0)
return ans;
// If kth node doesnt exists
// then break from the while loop
else
break;
}
q.push(NULL);
}
else {
while (temp) {
// If the required kth node
// has been found then return the answer
if (k == 0)
return ans;
k--;
// Update the value of variable ans
// each time
ans = temp->data;
if (temp->left)
q.push(temp->left);
temp = temp->right;
}
}
}
// If kth node doesnt exists then
// return -1
return -1;
}
// Driver Code
int main()
{
Node* root = newNode(8);
root->left = newNode(3);
root->right = newNode(10);
root->left->left = newNode(1);
root->left->right = newNode(6);
root->right->right = newNode(14);
root->right->right->left = newNode(13);
root->left->right->left = newNode(4);
root->left->right->right = newNode(7);
int k = 9;
cout << diagonalPrint(root, k);
return 0;
} Java
// Java program to print kth node
// in the diagonal traversal of a binary tree
import java.util.*;
class GFG
{
// A binary tree node has data, pointer to left
//child and a pointer to right child
static class Node
{
int data;
Node left, right;
};
// Helper function that allocates a new node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return (node);
}
// Iterative function to print kth node
// in diagonal traversal of binary tree
static int diagonalPrint(Node root, int k)
{
// Base cases
if (root == null || k == 0)
return -1;
int ans = -1;
Queue q = new LinkedList();
// add root node
q.add(root);
// add delimiter null
q.add(null);
while (q.size() > 0)
{
Node temp = q.peek();
q.remove();
if (temp == null)
{
if (q.size() == 0)
{
// If kth node exists then return
// the answer
if (k == 0)
return ans;
// If kth node doesnt exists
// then break from the while loop
else
break;
}
q.add(null);
}
else {
while (temp != null)
{
// If the required kth node
// has been found then return the answer
if (k == 0)
return ans;
k--;
// Update the value of variable ans
// each time
ans = temp.data;
if (temp.left!=null)
q.add(temp.left);
temp = temp.right;
}
}
}
// If kth node doesnt exists then
// return -1
return -1;
}
// Driver Code
public static void main(String args[])
{
Node root = newNode(8);
root.left = newNode(3);
root.right = newNode(10);
root.left.left = newNode(1);
root.left.right = newNode(6);
root.right.right = newNode(14);
root.right.right.left = newNode(13);
root.left.right.left = newNode(4);
root.left.right.right = newNode(7);
int k = 9;
System.out.println( diagonalPrint(root, k));
}
}
// This code is contributed by Arnab Kundu Python3
# Python program to print kth node
# in the diagonal traversal of a binary tree
# Linked List node
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Helper function that allocates a new node
def newNode(data) :
node = Node(0)
node.data = data
node.left = node.right = None
return (node)
# Iterative function to print kth node
# in diagonal traversal of binary tree
def diagonalPrint( root, k) :
# Base cases
if (root == None or k == 0) :
return -1
ans = -1
q = []
# append root node
q.append(root)
# append delimiter None
q.append(None)
while (len(q) > 0):
temp = q[0]
q.pop(0)
if (temp == None):
if (len(q) == 0) :
# If kth node exists then return
# the answer
if (k == 0) :
return ans
# If kth node doesnt exists
# then break from the while loop
else:
break
q.append(None)
else :
while (temp != None):
# If the required kth node
# has been found then return the answer
if (k == 0) :
return ans
k = k - 1
# Update the value of variable ans
# each time
ans = temp.data
if (temp.left != None):
q.append(temp.left)
temp = temp.right
# If kth node doesnt exists then
# return -1
return -1
# Driver Code
root = newNode(8)
root.left = newNode(3)
root.right = newNode(10)
root.left.left = newNode(1)
root.left.right = newNode(6)
root.right.right = newNode(14)
root.right.right.left = newNode(13)
root.left.right.left = newNode(4)
root.left.right.right = newNode(7)
k = 9
print( diagonalPrint(root, k))
# This code is contributed by Arnab KunduC#
// C# program to print kth node
// in the diagonal traversal of a binary tree
using System;
using System.Collections.Generic;
class GFG
{
// A binary tree node has data, pointer to left
//child and a pointer to right child
public class Node
{
public int data;
public Node left, right;
};
// Helper function that allocates a new node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return (node);
}
// Iterative function to print kth node
// in diagonal traversal of binary tree
static int diagonalPrint(Node root, int k)
{
// Base cases
if (root == null || k == 0)
return -1;
int ans = -1;
Queue q = new Queue();
// Enqueue root node
q.Enqueue(root);
// Enqueue delimiter null
q.Enqueue(null);
while (q.Count > 0)
{
Node temp = q.Peek();
q.Dequeue();
if (temp == null)
{
if (q.Count == 0)
{
// If kth node exists then return
// the answer
if (k == 0)
return ans;
// If kth node doesnt exists
// then break from the while loop
else
break;
}
q.Enqueue(null);
}
else
{
while (temp != null)
{
// If the required kth node
// has been found then return the answer
if (k == 0)
return ans;
k--;
// Update the value of variable ans
// each time
ans = temp.data;
if (temp.left!=null)
q.Enqueue(temp.left);
temp = temp.right;
}
}
}
// If kth node doesnt exists then
// return -1
return -1;
}
// Driver Code
public static void Main(String []args)
{
Node root = newNode(8);
root.left = newNode(3);
root.right = newNode(10);
root.left.left = newNode(1);
root.left.right = newNode(6);
root.right.right = newNode(14);
root.right.right.left = newNode(13);
root.left.right.left = newNode(4);
root.left.right.right = newNode(7);
int k = 9;
Console.WriteLine( diagonalPrint(root, k));
}
}
/* This code is contributed by PrinciRaj1992 */ Javascript
输出：
4时间复杂度：O(N)，其中 N 是二叉树中节点的总数。
辅助空间： O(N)