Java中创建二叉树并实现三种遍历方式

1、二叉树的创建

public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}

public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
if (pre == null || in == null) {
return null;
}
HashMap<Integer, Integer> hashMap = new HashMap<Integer, Integer>();
for (int i = 0; i < in.length; i++) {
hashMap.put(in[i], i);
}
return preIn(pre, 0, pre.length - 1, in, 0, in.length - 1, hashMap);
}

// 根据前序和中序还原二叉树
public TreeNode preIn(int[] p, int pi, int pj, int[] n, int ni, int nj,
HashMap<Integer, Integer> map) {
if (pi > pj) {
return null;
}
TreeNode head = new TreeNode(p[pi]);
int index = map.get(p[pi]);
// 根据前序遍历知道根节点，根据中序遍历知道根左边的为左子树，右边的为右子树
head.left = preIn(p, pi + 1, pi + index - ni, n, ni, index - 1, map);
head.right = preIn(p, pi + index - ni + 1, pj, n, index + 1, nj, map);
System.out
.println("head =" + (head != null ? head.val : " ")
+ " head.left ="
+ (head.left != null ? head.left.val : " ")
+ " head.right ="
+ (head.right != null ? head.right.val : " "));
return head;
}

2、前序遍历

// 根左右
public void preorderTraversa(TreeNode node) {
if (node == null) {
return;
}
System.out.print(node.val + " ");
preorderTraversa(node.left);
preorderTraversa(node.right);
}

3、中序遍历

// 左根右
public void inorderTraversa(TreeNode node) {
if (node == null) {
return;
}
inorderTraversa(node.left);
System.out.print(node.val + " ");
inorderTraversa(node.right);
}

4、后序遍历

// 左右根
public void postorderTraversa(TreeNode node) {
if (node == null) {
return;
}
postorderTraversa(node.left);
postorderTraversa(node.right);
System.out.print(node.val + " ");
}

5、具体调用代码如下

import java.util.HashMap;

public class BinaryTree {

public static void main(String[] args) {
int pre[] = { 1, 2, 4, 7, 3, 5, 6, 8 };
int in[] = { 4, 7, 2, 1, 5, 3, 8, 6 };
BinaryTree binaryTree = new BinaryTree();
// 利用前序和中序，创建二叉树
TreeNode treeNode = binaryTree.reConstructBinaryTree(pre, in);
// 前序遍历二叉树
binaryTree.preorderTraversa(treeNode);
System.out.println();
// 中序遍历
binaryTree.inorderTraversa(treeNode);
System.out.println();
// 后序遍历
binaryTree.postorderTraversa(treeNode);

}

// 左右根
public void postorderTraversa(TreeNode node) {
if (node == null) {
return;
}
postorderTraversa(node.left);
postorderTraversa(node.right);
System.out.print(node.val + " ");
}

// 根左右
public void preorderTraversa(TreeNode node) {
if (node == null) {
return;
}
System.out.print(node.val + " ");
preorderTraversa(node.left);
preorderTraversa(node.right);
}

// 左根右
public void inorderTraversa(TreeNode node) {
if (node == null) {
return;
}
inorderTraversa(node.left);
System.out.print(node.val + " ");
inorderTraversa(node.right);
}

public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
if (pre == null || in == null) {
return null;
}
HashMap<Integer, Integer> hashMap = new HashMap<Integer, Integer>();
for (int i = 0; i < in.length; i++) {
hashMap.put(in[i], i);
}
return preIn(pre, 0, pre.length - 1, in, 0, in.length - 1, hashMap);
}

// 根据前序和中序还原二叉树
public TreeNode preIn(int[] p, int pi, int pj, int[] n, int ni, int nj,
HashMap<Integer, Integer> map) {
if (pi > pj) {
return null;
}
TreeNode head = new TreeNode(p[pi]);
int index = map.get(p[pi]);
// 根据前序遍历知道根节点，根据中序遍历知道根左边的为左子树，右边的为右子树
head.left = preIn(p, pi + 1, pi + index - ni, n, ni, index - 1, map);
head.right = preIn(p, pi + index - ni + 1, pj, n, index + 1, nj, map);
System.out
.println("head =" + (head != null ? head.val : " ")
+ " head.left ="
+ (head.left != null ? head.left.val : " ")
+ " head.right ="
+ (head.right != null ? head.right.val : " "));
return head;
}
}