二叉树的遍历 [Java]
package cn.com.dom4j.adt.tree;import java.util.*;
public class TreeUtil {
/**
* 创建一个二叉树, 并返回其根节点 (从根节点开始每层从左至右依次填充)
*
* @param array 可变参数, 用来作为二叉树节点
* @param <AnyType> 泛型参数
* @return 根节点
*/
@SafeVarargs
public static <AnyType> BinaryNode<AnyType> createBinaryTree(AnyType... array) {
List<BinaryNode<AnyType>> nodeList = new ArrayList<>();
for (AnyType a : array) {
nodeList.add(new BinaryNode<>(a));
}
int length = array.length;
for (int i = 0; i < length / 2; i++) {
// 左孩子
nodeList.get(i).setLeft(nodeList.get(i * 2 + 1));
if (i != length / 2 - 1 || length % 2 == 1) {
// 右孩子
nodeList.get(i).setRight(nodeList.get(i * 2 + 2));
}
}
return nodeList.get(0);
}
/**
* 访问节点 (打印节点)
*/
public static <AnyType> void printNodeValue(BinaryNode<AnyType> node) {
System.out.print(node.getValue() + " ");
}
/**
* 层序遍历
*/
public static <AnyType> void layerOrder(BinaryNode<AnyType> root) {
Queue<BinaryNode<AnyType>> queue = new LinkedList<>();
// 从根节点入队列
if (root != null)
queue.offer(root);
while (!queue.isEmpty()) {
// 取出队列首节点
BinaryNode<AnyType> node = queue.poll();
printNodeValue(node);
if (node.getLeft() != null) {
// 左节点入队列
queue.offer(node.getLeft());
}
if (node.getRight() != null) {
// 右节点入队列
queue.offer(node.getRight());
}
}
}
/**
* 前序遍历 递归实现
*/
public static <AnyType> void preOrderByRecursion(BinaryNode<AnyType> node) {
if (node == null)
return;
System.out.print(node.getValue() + " ");
preOrderByRecursion(node.getLeft());
preOrderByRecursion(node.getRight());
}
/**
* 中序遍历 递归实现
*/
public static <AnyType> void inOrderByRecursion(BinaryNode<AnyType> node) {
if (node == null)
return;
inOrderByRecursion(node.getLeft());
printNodeValue(node);
inOrderByRecursion(node.getRight());
}
/**
* 后续遍历 递归实现
*/
public static <AnyType> void postOrderByRecursion(BinaryNode<AnyType> node) {
if (node == null)
return;
postOrderByRecursion(node.getLeft());
postOrderByRecursion(node.getRight());
printNodeValue(node);
}
/**
* 前序遍历, 非递归实现
* 1. 先入栈根节点, 打印其值, 再先后入栈右节点,左节点
* 2. 出栈左节点, 打印其值, 再入栈该左节点的右节点,左节点, 直到遍历完该左节点所在子树
* 3. 再出栈右节点, 打印其值, 再入栈该右节点的右节点,左节点, 直到遍历完该右节点所在子树
*/
public static <AnyType> void preOrderByNoRecursion(BinaryNode<AnyType> node) {
Stack<BinaryNode<AnyType>> stack = new Stack<>();
if (node != null) {
stack.push(node);
}
while (!stack.empty()) {
BinaryNode<AnyType> n = stack.pop();
printNodeValue(n);
if (n.getRight() != null) {
stack.push(n.getRight());
}
if (n.getLeft() != null) {
stack.push(n.getLeft());
}
}
}
/**
* 中序遍历 非递归实现
*/
public static <AnyType> void inOrderByNoRecursion(BinaryNode<AnyType> node) {
Stack<BinaryNode<AnyType>> stack = new Stack<>();
while (node != null || !stack.empty()) {
while (node != null) {
stack.push(node);
node = node.getLeft();
}
if (!stack.empty()) {
node = stack.pop();
printNodeValue(node);
node = node.getRight();
}
}
}
}
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