用父指针查找二叉树的右兄弟
给定具有父指针的二叉树，找到给定节点的右兄弟（将给出指向该节点的指针），如果不存在则返回 null。在 O(1) 空间和 O(n) 时间内完成吗？
例子：
1
/ \
2 3
/ \ \
4 6 5
/ \ \
7 9 8
/ \
10 12
Input : Given above tree with parent pointer and node 10
Output : 12 方法：这个想法是找出最近的祖先的第一个右孩子，它既不是当前节点也不是当前节点的父节点，在上升的同时跟踪那些级别。然后，遍历该节点的第一个左子节点，如果左边不存在则为右子节点，如果级别变为 0，则这是给定节点的下一个右兄弟节点。
在上面的例子中，如果给定的节点是 7，我们最终会得到 6 来找到没有任何子节点的正确子节点。
在这种情况下，我们需要递归调用具有当前级别的右兄弟，以便我们达到 8。
C++
// C program to print right sibling of a node
#include
// A Binary Tree Node
struct Node {
int data;
Node *left, *right, *parent;
};
// A utility function to create a new Binary
// Tree Node
Node* newNode(int item, Node* parent)
{
Node* temp = new Node;
temp->data = item;
temp->left = temp->right = NULL;
temp->parent = parent;
return temp;
}
// Method to find right sibling
Node* findRightSibling(Node* node, int level)
{
if (node == NULL || node->parent == NULL)
return NULL;
// GET Parent pointer whose right child is not
// a parent or itself of this node. There might
// be case when parent has no right child, but,
// current node is left child of the parent
// (second condition is for that).
while (node->parent->right == node
|| (node->parent->right == NULL
&& node->parent->left == node)) {
if (node->parent == NULL
|| node->parent->parent == NULL)
return NULL;
node = node->parent;
level--;
}
// Move to the required child, where right sibling
// can be present
node = node->parent->right;
if (node == NULL)
return NULL;
// find right sibling in the given subtree(from current
// node), when level will be 0
while (level < 0) {
// Iterate through subtree
if (node->left != NULL)
node = node->left;
else if (node->right != NULL)
node = node->right;
else
// if no child are there, we cannot have right
// sibling in this path
break;
level++;
}
if (level == 0)
return node;
// This is the case when we reach 9 node in the tree,
// where we need to again recursively find the right
// sibling
return findRightSibling(node, level);
}
// Driver Program to test above functions
int main()
{
Node* root = newNode(1, NULL);
root->left = newNode(2, root);
root->right = newNode(3, root);
root->left->left = newNode(4, root->left);
root->left->right = newNode(6, root->left);
root->left->left->left = newNode(7, root->left->left);
root->left->left->left->left = newNode(10, root->left->left->left);
root->left->right->right = newNode(9, root->left->right);
root->right->right = newNode(5, root->right);
root->right->right->right = newNode(8, root->right->right);
root->right->right->right->right = newNode(12, root->right->right->right);
// passing 10
Node* res = findRightSibling(root->left->left->left->left, 0);
if (res == NULL)
printf("No right sibling");
else
printf("%d", res->data);
return 0;
} Java
// Java program to print right sibling of a node
public class Right_Sibling {
// A Binary Tree Node
static class Node {
int data;
Node left, right, parent;
// Constructor
public Node(int data, Node parent)
{
this.data = data;
left = null;
right = null;
this.parent = parent;
}
};
// Method to find right sibling
static Node findRightSibling(Node node, int level)
{
if (node == null || node.parent == null)
return null;
// GET Parent pointer whose right child is not
// a parent or itself of this node. There might
// be case when parent has no right child, but,
// current node is left child of the parent
// (second condition is for that).
while (node.parent.right == node
|| (node.parent.right == null
&& node.parent.left == node)) {
if (node.parent == null)
return null;
node = node.parent;
level--;
}
// Move to the required child, where right sibling
// can be present
node = node.parent.right;
// find right sibling in the given subtree(from current
// node), when level will be 0
while (level < 0) {
// Iterate through subtree
if (node.left != null)
node = node.left;
else if (node.right != null)
node = node.right;
else
// if no child are there, we cannot have right
// sibling in this path
break;
level++;
}
if (level == 0)
return node;
// This is the case when we reach 9 node in the tree,
// where we need to again recursively find the right
// sibling
return findRightSibling(node, level);
}
// Driver Program to test above functions
public static void main(String args[])
{
Node root = new Node(1, null);
root.left = new Node(2, root);
root.right = new Node(3, root);
root.left.left = new Node(4, root.left);
root.left.right = new Node(6, root.left);
root.left.left.left = new Node(7, root.left.left);
root.left.left.left.left = new Node(10, root.left.left.left);
root.left.right.right = new Node(9, root.left.right);
root.right.right = new Node(5, root.right);
root.right.right.right = new Node(8, root.right.right);
root.right.right.right.right = new Node(12, root.right.right.right);
// passing 10
System.out.println(findRightSibling(root.left.left.left.left, 0).data);
}
}
// This code is contributed by Sumit GhoshPython3
# Python3 program to print right sibling
# of a node
# A class to create a new Binary
# Tree Node
class newNode:
def __init__(self, item, parent):
self.data = item
self.left = self.right = None
self.parent = parent
# Method to find right sibling
def findRightSibling(node, level):
if (node == None or node.parent == None):
return None
# GET Parent pointer whose right child is not
# a parent or itself of this node. There might
# be case when parent has no right child, but,
# current node is left child of the parent
# (second condition is for that).
while (node.parent.right == node or
(node.parent.right == None and
node.parent.left == node)):
if (node.parent == None):
return None
node = node.parent
level -= 1
# Move to the required child, where
# right sibling can be present
node = node.parent.right
# find right sibling in the given subtree
# (from current node), when level will be 0
while (level < 0):
# Iterate through subtree
if (node.left != None):
node = node.left
else if (node.right != None):
node = node.right
else:
# if no child are there, we cannot
# have right sibling in this path
break
level += 1
if (level == 0):
return node
# This is the case when we reach 9 node
# in the tree, where we need to again
# recursively find the right sibling
return findRightSibling(node, level)
# Driver Code
if __name__ == '__main__':
root = newNode(1, None)
root.left = newNode(2, root)
root.right = newNode(3, root)
root.left.left = newNode(4, root.left)
root.left.right = newNode(6, root.left)
root.left.left.left = newNode(7, root.left.left)
root.left.left.left.left = newNode(10, root.left.left.left)
root.left.right.right = newNode(9, root.left.right)
root.right.right = newNode(5, root.right)
root.right.right.right = newNode(8, root.right.right)
root.right.right.right.right = newNode(12, root.right.right.right)
# passing 10
res = findRightSibling(root.left.left.left.left, 0)
if (res == None):
print("No right sibling")
else:
print(res.data)
# This code is contributed by PranchalKC#
using System;
// C# program to print right sibling of a node
public class Right_Sibling {
// A Binary Tree Node
public class Node {
public int data;
public Node left, right, parent;
// Constructor
public Node(int data, Node parent)
{
this.data = data;
left = null;
right = null;
this.parent = parent;
}
}
// Method to find right sibling
public static Node findRightSibling(Node node, int level)
{
if (node == null || node.parent == null) {
return null;
}
// GET Parent pointer whose right child is not
// a parent or itself of this node. There might
// be case when parent has no right child, but,
// current node is left child of the parent
// (second condition is for that).
while (node.parent.right == node
|| (node.parent.right == null
&& node.parent.left == node)) {
if (node.parent == null
|| node.parent.parent == null) {
return null;
}
node = node.parent;
level--;
}
// Move to the required child, where right sibling
// can be present
node = node.parent.right;
// find right sibling in the given subtree(from current
// node), when level will be 0
while (level < 0) {
// Iterate through subtree
if (node.left != null) {
node = node.left;
}
else if (node.right != null) {
node = node.right;
}
else {
// if no child are there, we cannot have right
// sibling in this path
break;
}
level++;
}
if (level == 0) {
return node;
}
// This is the case when we reach 9 node in the tree,
// where we need to again recursively find the right
// sibling
return findRightSibling(node, level);
}
// Driver Program to test above functions
public static void Main(string[] args)
{
Node root = new Node(1, null);
root.left = new Node(2, root);
root.right = new Node(3, root);
root.left.left = new Node(4, root.left);
root.left.right = new Node(6, root.left);
root.left.left.left = new Node(7, root.left.left);
root.left.left.left.left = new Node(10, root.left.left.left);
root.left.right.right = new Node(9, root.left.right);
root.right.right = new Node(5, root.right);
root.right.right.right = new Node(8, root.right.right);
root.right.right.right.right = new Node(12, root.right.right.right);
// passing 10
Console.WriteLine(findRightSibling(root.left.left.left.left, 0).data);
}
}
// This code is contributed by Shrikant13Javascript
输出：
12时间复杂度： O(N)
辅助空间： O(1)