java使用递归及迭代方式实现前序遍历 中序遍历 后序遍历 以及实现层序遍历
本文为博主原创，转载请注明出处：
目录：
一。快速理解前序，中序，后序遍历的区别
二。使用递归的方式实现前序，中序，后序遍历
三。 使用迭代的方式实现前序 中序 后序遍历
四。层序遍历
一。快速理解前序，中序，后序遍历的区别
前序遍历：根左右（根在前，从左往右，一棵树的根永远在左子树前面，左子树又永远在右子树前面 ）
中序遍历：左根右（根在中，从左往右，一棵树的左子树永远在根前面，根永远在右子树前面）
后序遍历：左右根（根在后，从左往右，一棵树的左子树永远在右子树前面，右子树永远在根前面）
参考一张图可以快速理解三种遍历的顺序
二。使用递归的方式实现前序，中序，后序遍历
package com.example.test.tree;
public class TreeNode {
// 树的根结点值
int val;
// 根节点对应的左节点
TreeNode left = null;
// 根节点对应的右节点
TreeNode right = null;
public TreeNode(int val){
this.val = val;
}
/**
* 树的前序便利：根节点--〉左节点---〉右节点
* @param root
*/
public static void preOrder(TreeNode root) {
if (root == null) {
return;
}
System.out.print(root.val + " ");
preOrder(root.left);
preOrder(root.right);
}
/**
* 树的中序遍历 ： 左节点--〉中节点 --〉 右节点
* @param root
*/
public static void middleOrder(TreeNode root){
if (root ==null){return;}
middleOrder(root.left);
System.out.print(root.val+ " ");
middleOrder(root.right);
}
/**
* 树的后序遍历： 左节点--〉右节点 --> 根节点
* @param root
*/
public static void afterOrder(TreeNode root){
if (root == null){return;}
afterOrder(root.left);
afterOrder(root.right);
System.out.print(root.val+ " ");
}
public static void main(String[] args) {
TreeNode root = new TreeNode(20);
TreeNode node1 = new TreeNode(4);
TreeNode node2 = new TreeNode(5);
TreeNode node3 = new TreeNode(8);
TreeNode node4 = new TreeNode(2);
TreeNode node5 = new TreeNode(7);
TreeNode node6 = new TreeNode(12);
TreeNode node7 = new TreeNode(3);
TreeNode node8 = new TreeNode(9);
root.left = node1;
root.right = node2;
node1.left = node3;
node1.right = node4;
node2.left = node5;
node2.right = node6;
node3.left = node7;
node4.right = node8;
preOrder(root);
System.out.println();
middleOrder(root);
System.out.println();
afterOrder(root);
}
}
三。 使用迭代的方式实现前序 中序 后序遍历
package com.example.test.tree;
import java.util.Stack;
public class TreeNode {
// 树的根结点值
int val;
// 根节点对应的左节点
TreeNode left = null;
// 根节点对应的右节点
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
/**
* 前序遍历：使用stack 记录递归路径，须保证左子节点先出栈
*
* @param root
*/
public static void preOrder2(TreeNode root) {
if (root != null) {
Stack<TreeNode> stack = new Stack<>();
stack.add(root);
while (!stack.isEmpty()) {
root = stack.pop();
if (root != null) {
System.out.print(root.val + " ");
stack.push(root.right);
stack.push(root.left);
}
}
}
}
/**
* 中序遍历：将左子节点入栈，出栈打印值，然后添加右子节点
*
* @param root
*/
public static void middleOrder2(TreeNode root) {
if (root != null) {
Stack<TreeNode> stack = new Stack<>();
while (!stack.isEmpty() || root != null)
if (root != null) {
stack.push(root);
root = root.left;
} else {
root = stack.pop();
System.out.print(root.val + " ");
root = root.right;
}
}
}
/**
* 后序遍历
* @param root
*/
public void afterOrder2(TreeNode root) {
TreeNode cur, pre = null;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.empty()) {
cur = stack.peek();
if ((cur.left == null && cur.right == null) || (pre != null && (pre == cur.left || pre == cur.right))) {
System.out.print(cur.val + "->");
stack.pop();
pre = cur;
} else {
if (cur.right != null)
stack.push(cur.right);
if (cur.left != null)
stack.push(cur.left);
}
}
}
public static void main(String[] args) {
TreeNode root = new TreeNode(20);
TreeNode node1 = new TreeNode(4);
TreeNode node2 = new TreeNode(5);
TreeNode node3 = new TreeNode(8);
TreeNode node4 = new TreeNode(2);
TreeNode node5 = new TreeNode(7);
TreeNode node6 = new TreeNode(12);
TreeNode node7 = new TreeNode(3);
TreeNode node8 = new TreeNode(9);
root.left = node1;
root.right = node2;
node1.left = node3;
node1.right = node4;
node2.left = node5;
node2.right = node6;
node3.left = node7;
node4.right = node8;
preOrder2(root);
System.out.println();
middleOrder2(root);
System.out.println();
afterOrder2(root);
}
}
四。层序遍历
/**
* 层序遍历
* @param root
*/
public static void levelOrder(TreeNode root) {
if (root == null) {
return;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
System.out.print(node.val + "->");
if (node.left != null) {
queue.add(node.left);
}
if (node.right != null) {
queue.add(node.right);
}
}
}
