连接层序兄弟结点
连接层序兄弟结点
题目
给定一棵二叉树,将每个节点与其层次顺序的后继节点连接起来。每个级别的最后一个节点应该指向一个空节点。
解答
import java.util.*;
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode next;
TreeNode(int x) {
val = x;
left = right = next = null;
}
//打印节点 level order traversal using 'next' pointer
public void printLevelOrder() {
TreeNode nextLevelRoot = this;
while (nextLevelRoot != null) {
TreeNode current = nextLevelRoot;
nextLevelRoot = null;
while (current != null) {
System.out.print(current.val + " ");
if (nextLevelRoot == null) {
if (current.left != null)
nextLevelRoot = current.left;
else if (current.right != null)
nextLevelRoot = current.right;
}
current = current.next;
}
System.out.println();
}
}
};
class ConnectLevelOrderSiblings {
public static void connect(TreeNode root) {
if (root == null)
return;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
TreeNode previousNode = null;
int levelSize = queue.size();
//遍历这一层的所有节点
for (int i = 0; i < levelSize; i++) {
TreeNode currentNode = queue.poll();
if (previousNode != null)
previousNode.next = currentNode;
previousNode = currentNode;
//插入当前节点的子节点
if (currentNode.left != null)
queue.offer(currentNode.left);
if (currentNode.right != null)
queue.offer(currentNode.right);
}
}
}
public static void main(String[] args) {
TreeNode root = new TreeNode(12);
root.left = new TreeNode(7);
root.right = new TreeNode(1);
root.left.left = new TreeNode(9);
root.right.left = new TreeNode(10);
root.right.right = new TreeNode(5);
ConnectLevelOrderSiblings.connect(root);
System.out.println("Level order traversal using 'next' pointer: ");
root.printLevelOrder();
}
}
class TreeNode {
constructor(x) {
this.val = x;
this.left = null;
this.right = null;
this.next = null;
}
//打印节点 level order traversal using 'next' pointer
printLevelOrder() {
let nextLevelRoot = this;
while (nextLevelRoot !== null) {
let current = nextLevelRoot;
nextLevelRoot = null;
while (current !== null) {
console.log(current.val + " ");
if (nextLevelRoot === null) {
if (current.left !== null)
nextLevelRoot = current.left;
else if (current.right !== null)
nextLevelRoot = current.right;
}
current = current.next;
}
console.log();
}
}
}
class ConnectLevelOrderSiblings {
static connect(root) {
if (root === null)
return;
let queue = [];
queue.push(root);
while (queue.length > 0) {
let previousNode = null;
let levelSize = queue.length;
//遍历这一层的所有节点
for (let i = 0; i < levelSize; i++) {
let currentNode = queue.shift();
if (previousNode !== null)
previousNode.next = currentNode;
previousNode = currentNode;
//插入当前节点的子节点
if (currentNode.left !== null)
queue.push(currentNode.left);
if (currentNode.right !== null)
queue.push(currentNode.right);
}
}
}
static main() {
let root = new TreeNode(12);
root.left = new TreeNode(7);
root.right = new TreeNode(1);
root.left.left = new TreeNode(9);
root.right.left = new TreeNode(10);
root.right.right = new TreeNode(5);
ConnectLevelOrderSiblings.connect(root);
console.log("Level order traversal using 'next' pointer: ");
root.printLevelOrder();
}
}
ConnectLevelOrderSiblings.main();
from collections import deque
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
self.next = None
# 打印节点 level order traversal using 'next' pointer
def printLevelOrder(self):
nextLevelRoot = self
while nextLevelRoot:
current = nextLevelRoot
nextLevelRoot = None
while current:
print(current.val, end=" ")
if not nextLevelRoot:
if current.left:
nextLevelRoot = current.left
elif current.right:
nextLevelRoot = current.right
current = current.next
print()
class ConnectLevelOrderSiblings:
@staticmethod
def connect(root):
if not root:
return
queue = deque()
queue.append(root)
while queue:
previousNode = None
levelSize = len(queue)
# 遍历这一层的所有节点
for _ in range(levelSize):
currentNode = queue.popleft()
if previousNode:
previousNode.next = currentNode
previousNode = currentNode
# 插入当前节点的子节点
if currentNode.left:
queue.append(currentNode.left)
if currentNode.right:
queue.append(currentNode.right)
@staticmethod
def main():
root = TreeNode(12)
root.left = TreeNode(7)
root.right = TreeNode(1)
root.left.left = TreeNode(9)
root.right.left = TreeNode(10)
root.right.right = TreeNode(5)
ConnectLevelOrderSiblings.connect(root)
print("Level order traversal using 'next' pointer: ")
root.printLevelOrder()
ConnectLevelOrderSiblings.main()
package main
import (
"fmt"
)
type TreeNode struct {
val int
left *TreeNode
right *TreeNode
next *TreeNode
}
//打印节点 level order traversal using 'next' pointer
func (node *TreeNode) printLevelOrder() {
nextLevelRoot := node
for nextLevelRoot != nil {
current := nextLevelRoot
nextLevelRoot = nil
for current != nil {
fmt.Printf("%d ", current.val)
if nextLevelRoot == nil {
if current.left != nil {
nextLevelRoot = current.left
} else if current.right != nil {
nextLevelRoot = current.right
}
}
current = current.next
}
fmt.Println()
}
}
type ConnectLevelOrderSiblings struct{}
func (c *ConnectLevelOrderSiblings) connect(root *TreeNode) {
if root == nil {
return
}
queue := []*TreeNode{root}
for len(queue) > 0 {
previousNode := (*TreeNode)(nil)
levelSize := len(queue)
//遍历这一层的所有节点
for i := 0; i < levelSize; i++ {
currentNode := queue[0]
queue = queue[1:]
if previousNode != nil {
previousNode.next = currentNode
}
previousNode = currentNode
//插入当前节点的子节点
if currentNode.left != nil {
queue = append(queue, currentNode.left)
}
if currentNode.right != nil {
queue = append(queue, currentNode.right)
}
}
}
}
func main() {
root := &TreeNode{val: 12}
root.left = &TreeNode{val: 7}
root.right = &TreeNode{val: 1}
root.left.left = &TreeNode{val: 9}
root.right.left = &TreeNode{val: 10}
root.right.right = &TreeNode{val: 5}
connectSiblings := &ConnectLevelOrderSiblings{}
connectSiblings.connect(root)
fmt.Println("Level order traversal using 'next' pointer:")
root.printLevelOrder()
}