节点访问的次序,忽略打印行为
如果将打印安排在同个数字第一次被访问时,即先序遍历
第二次即中序遍历
第三次即后序遍历
现二叉树的先序、中序、后序遍历，包括递归方式和非递归
方式
二叉树结构定义
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
递归
先序遍历
/**
* 先序遍历
*
* @param head
*/
public static void preOrderRecur(Node head) {
if (head == null) {
return;
}
System.out.print(head.value + " ");
preOrderRecur(head.left);
preOrderRecur(head.right);
}
中序遍历
/**
* 中序遍历
*
* @param head
*/
public static void inOrderRecur(Node head) {
if (head == null) {
return;
}
inOrderRecur(head.left);
System.out.print(head.value + " ");
in
后序遍历
/**
* 后序遍历
*
* @param head
*/
public static void posOrderRecur(Node head) {
if (head == null) {
return;
}
posOrderRecur(head.left);
posOrderRecur(head.right);
System.out.print(head.value + " ");
}
2 非递归
2.1 先序遍历非递归
public static void preOrderUnRecur(Node head) {
System.out.print("pre-order: ");
if (head != null) {
//构造栈
Stack<Node> stack =new Stack<>();
//压栈以当前结点为头,变成先序的逆序
stack.add(head);
while (!stack.isEmpty()) {
//弹栈
head = stack.pop();
System.out.print(head.value + " ");
/**
* 有右孩子就先压,否则压左
*/
//右孩子非空时
if (head.right != null) {
//压栈右孩子
stack.push(head.right);
}
//左孩子非空时
if (head.left != null) {
//压栈左孩子
stack.push(head.left);
}
}
}
System.out.println();
}
2.2 中序遍历非递归
/**
* 中序遍历的非递归版本
*
* @param head
*/
public static void inOrderUnRecur(Node head) {
System.out.print("in-order: ");
if (head != null) {
//构造栈
Stack<Node> stack =new Stack<>();
//当前结点为空,从栈弹一个打印,当前结点向右跑
// 非空,则压栈,向左跑
while (!stack.isEmpty() || head != null) {
//会把当前结点的左孩子全都压栈!
if (head != null) {
stack.push(head);
head = head.left;
} else {
head = stack.pop();
System.out.print(head.value + " ");
head = head.right;
}
}
}
System.out.println();
}
整棵树都被左边界分解,然后栈的逆序打印即可
再怎么解释都需要自己的代码理解!!!中序遍历非递归也是最难理解的!
2.3 后序遍历非递归
即要实现左右中,逆序中左右其实根据先序遍历非递归改变左右压栈顺序即可,然后该打印时不打印,而是放在辅助栈中,就成了左右中顺序,打印之!
/**
* 后序遍历的非递归版本1
*
* @param head
*/
public static void posOrderUnRecur1(Node head) {
System.out.print("pos-order: ");
if (head != null) {
Stack<Node> s1 =new Stack<>();
//辅助栈
Stack<Node> s2 =new Stack<>();
s1.push(head);
while (!s1.isEmpty()) {
head = s1.pop();
//打印时,不让其打印,而是放在辅助栈
s2.push(head);
//先压左
if (head.left != null) {
s1.push(head.left);
}
//再压右
if (head.right != null) {
s1.push(head.right);
}
}
while (!s2.isEmpty()) {
//最后打印辅助栈即可!
System.out.print(s2.pop().value + " ");
}
}
System.out.println();
}
/**
* 后序遍历的非递归版本2
*
* @param h
*/
public static void posOrderUnRecur2(Node h) {
System.out.print("pos-order: ");
if (h != null) {
Stack<Node> stack =new Stack<>();
stack.push(h);
Node c = null;
while (!stack.isEmpty()) {
c = stack.peek();
if (c.left != null && h != c.left && h != c.right) {
stack.push(c.left);
} else if (c.right != null && h != c.right) {
stack.push(c.right);
} else {
System.out.print(stack.pop().value + " ");
h = c;
}
}
}
System.out.println();
}
3 小结
无论递归非递归,本质都是使用了栈ADT,由于二叉树本身只有向下的路径,所以需要有支持回去的路径,栈即为天之骄子!!!二叉树整体是向下的,当遍历到某一结点时需要回去时,就是刚刚好逆序的栈ADT!
福利函数,打印正确的二叉树,用于验证
/**
* @author Shusheng Shi
*/
public class PrintBinaryTree {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
public static void printTree(Node head) {
System.out.println("Binary Tree:");
printInOrder(head, 0, "H", 17);
System.out.println();
}
public static void printInOrder(Node head, int height, String to, int len) {
if (head == null) {
return;
}
printInOrder(head.right, height + 1, "v", len);
String val = to + head.value + to;
int lenM = val.length();
int lenL = (len - lenM) / 2;
int lenR = len - lenM - lenL;
val = getSpace(lenL) + val + getSpace(lenR);
System.out.println(getSpace(height * len) + val);
printInOrder(head.left, height + 1, "^", len);
}
public static String getSpace(int num) {
String space = " ";
StringBuffer buf = new StringBuffer("");
for (int i = 0; i < num; i++) {
buf.append(space);
}
return buf.toString();
}
public static void main(String[] args) {
Node head = new Node(1);
head.left = new Node(-222222222);
head.right = new Node(3);
head.left.left = new Node(Integer.MIN_VALUE);
head.right.left = new Node(55555555);
head.right.right = new Node(66);
head.left.left.right = new Node(777);
printTree(head);
head = new Node(1);
head.left = new Node(2);
head.right = new Node(3);
head.left.left = new Node(4);
head.right.left = new Node(5);
head.right.right = new Node(6);
head.left.left.right = new Node(7);
printTree(head);
head = new Node(1);
head.left = new Node(1);
head.right = new Node(1);
head.left.left = new Node(1);
head.right.left = new Node(1);
head.right.right = new Node(1);
head.left.left.right = new Node(1);
printTree(head);
}
}
4 实战coding
在二叉树中找到一个节点的后继节点
现在有一种新的二叉树节点类型如下：
public class Node {
public int value;
public Node left;
public Node right;
//可找到父结点,头结点指空
public Node parent;
public Node(int data) {
this.value = data;
}
}
该结构比普通二叉树节点结构多了一个指向父节点的parent指针。
假设有一 棵Node类型的节点组成的二叉树，树中每个节点的parent指针都正确地指向 自己的父节点，头节点的parent指向null。
只给一个在二叉树中的某个节点 node，请实现返回node的后继节点的函数。
在二叉树的中序遍历的序列中， node的下一个节点叫作node的后继节点
后继结点示例
前驱结点同理!不赘述
/**
* @author Shusheng Shi
*/
public class SuccessorNode {
public static class Node {
public int value;
public Node left;
public Node right;
public Node parent;
public Node(int data) {
this.value = data;
}
}
/**
* 如果是父节点的左孩子,就直接打印父节点
* 否则找父节点的父节点,看父节点是否为其父的左孩子
* 一路上行
* @param node
* @return
*/
public static Node getSuccessorNode(Node node) {
if (node == null) {
return node;
}
//当前节点右孩子非空,说明有右子树
if (node.right != null) {
//直接找右子树的最左结点
return getLeftMost(node.right);
//当前节点右孩子为空
} else {
//取得其父结点
Node parent = node.parent;
//当前结点为其父左孩子时退出循环,可以简单认为整棵树为空节点的左孩子
while (parent != null && parent.left != node) {
//否则两指针同时一路上行
node = parent;
parent = node.parent;
}
return parent;
}
}
/**
* 给定结点子树的最左结点
*
* @param node 某棵子树的头部
* @return
*/
public static Node getLeftMost(Node node) {
if (node == null) {
return node;
}
while (node.left != null) {
//一路左行即可
node = node.left;
}
return node;
}
public static void main(String[] args) {
Node head = new Node(6);
head.parent = null;
head.left = new Node(3);
head.left.parent = head;
head.left.left = new Node(1);
head.left.left.parent = head.left;
head.left.left.right = new Node(2);
head.left.left.right.parent = head.left.left;
head.left.right = new Node(4);
head.left.right.parent = head.left;
head.left.right.right = new Node(5);
head.left.right.right.parent = head.left.right;
head.right = new Node(9);
head.right.parent = head;
head.right.left = new Node(8);
head.right.left.parent = head.right;
head.right.left.left = new Node(7);
head.right.left.left.parent = head.right.left;
head.right.right = new Node(10);
head.right.right.parent = head.right;
Node test = head.left.left;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.left.left.right;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.left;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.left.right;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.left.right.right;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.right.left.left;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.right.left;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.right;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.right.right; // 10's next is null
System.out.println(test.value + " next: " + getSuccessorNode(test));
}
}
5序列化和反序列化
怎么序列化的就怎么反序列化
按层序列化示意
import java.util.LinkedList;
import java.util.Queue;
/**
* @author Shusheng Shi
*/
public class SerializeAndReconstructTree {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
public static String serialByPre(Node head) {
if (head == null) {
return "#!";
}
String res = head.value + "!";
res += serialByPre(head.left);
res += serialByPre(head.right);
return res;
}
public static Node reconByPreString(String preStr) {
String[] values = preStr.split("!");
//加入队列,方便依次出队
Queue<String> queue =new LinkedList<>();
for (int i = 0; i != values.length; i++) {
queue.offer(values[i]);
}
return reconPreOrder(queue);
}
/**
* 用给定队列建树
*
* @param queue
* @return
*/
public static Node reconPreOrder(Queue<String> queue) {
//出队一个
String value = queue.poll();
//空树
if (value.equals("#")) {
return null;
}
//非空时建立一个头结点,中-左-右顺序
Node head = new Node(Integer.valueOf(value));
//左子树递归
head.left = reconPreOrder(queue);
//右子树递归
head.right = reconPreOrder(queue);
return head;
}
/**
* 按层序列化
*
* @param head
* @return
*/
public static String serialByLevel(Node head) {
if (head == null) {
return "#!";
}
String res = head.value + "!";
Queue<Node> queue = new LinkedList<Node>();
queue.offer(head);
while (!queue.isEmpty()) {
head = queue.poll();
if (head.left != null) {
res += head.left.value + "!";
queue.offer(head.left);
} else {
res += "#!";
}
if (head.right != null) {
res += head.right.value + "!";
queue.offer(head.right);
} else {
res += "#!";
}
}
return res;
}
public static Node reconByLevelString(String levelStr) {
String[] values = levelStr.split("!");
int index = 0;
Node head = generateNodeByString(values[index++]);
Queue<Node> queue = new LinkedList<Node>();
if (head != null) {
queue.offer(head);
}
Node node = null;
while (!queue.isEmpty()) {
node = queue.poll();
node.left = generateNodeByString(values[index++]);
node.right = generateNodeByString(values[index++]);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
return head;
}
public static Node generateNodeByString(String val) {
if (val.equals("#")) {
return null;
}
return new Node(Integer.valueOf(val));
}
public static void printTree(Node head) {
System.out.println("Binary Tree:");
printInOrder(head, 0, "H", 17);
System.out.println();
}
public static void printInOrder(Node head, int height, String to, int len) {
if (head == null) {
return;
}
printInOrder(head.right, height + 1, "v", len);
String val = to + head.value + to;
int lenM = val.length();
int lenL = (len - lenM) / 2;
int lenR = len - lenM - lenL;
val = getSpace(lenL) + val + getSpace(lenR);
System.out.println(getSpace(height * len) + val);
printInOrder(head.left, height + 1, "^", len);
}
public static String getSpace(int num) {
String space = " ";
StringBuffer buf = new StringBuffer("");
for (int i = 0; i < num; i++) {
buf.append(space);
}
return buf.toString();
}
public static void main(String[] args) {
Node head = null;
printTree(head);
String pre = serialByPre(head);
System.out.println("serialize tree by pre-order: " + pre);
head = reconByPreString(pre);
System.out.print("reconstruct tree by pre-order, ");
printTree(head);
String level = serialByLevel(head);
System.out.println("serialize tree by level: " + level);
head = reconByLevelString(level);
System.out.print("reconstruct tree by level, ");
printTree(head);
System.out.println("====================================");
head = new Node(1);
printTree(head);
pre = serialByPre(head);
System.out.println("serialize tree by pre-order: " + pre);
head = reconByPreString(pre);
System.out.print("reconstruct tree by pre-order, ");
printTree(head);
level = serialByLevel(head);
System.out.println("serialize tree by level: " + level);
head = reconByLevelString(level);
System.out.print("reconstruct tree by level, ");
printTree(head);
System.out.println("====================================");
head = new Node(1);
head.left = new Node(2);
head.right = new Node(3);
head.left.left = new Node(4);
head.right.right = new Node(5);
printTree(head);
pre = serialByPre(head);
System.out.println("serialize tree by pre-order: " + pre);
head = reconByPreString(pre);
System.out.print("reconstruct tree by pre-order, ");
printTree(head);
level = serialByLevel(head);
System.out.println("serialize tree by level: " + level);
head = reconByLevelString(level);
System.out.print("reconstruct tree by level, ");
printTree(head);
System.out.println("====================================");
head = new Node(100);
head.left = new Node(21);
head.left.left = new Node(37);
head.right = new Node(-42);
head.right.left = new Node(0);
head.right.right = new Node(666);
printTree(head);
pre = serialByPre(head);
System.out.println("serialize tree by pre-order: " + pre);
head = reconByPreString(pre);
System.out.print("reconstruct tree by pre-order, ");
printTree(head);
level = serialByLevel(head);
System.out.println("serialize tree by level: " + level);
head = reconByLevelString(level);
System.out.print("reconstruct tree by level, ");
printTree(head);
System.out.println("====================================");
}
}
判断一棵二叉树是否是平衡二叉树
平衡二叉树（Self-balancing binary search tree）
又被称为AVL树（有别于AVL算法），且具有以下性质：它是一 棵空树或它的左右两个子树的高度差的绝对值不超过1，并且左右两个子树都是一棵平衡二叉树，同时，平衡二叉树必定是二叉搜索树，反之则不一定。平衡二叉树的常用实现方法有红黑树、AVL、替罪羊树、Treap、伸展树等。
解题思路
以每个结点为头的整棵树都要是平衡的
左子树平衡?
右子树平衡?
都平衡情况下,左子树高度?
右子树高度?
判断差值
package com.sss.class04;
/**
* @author Shusheng Shi
*/
public class IsBalancedTree {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
public static boolean isBalance(Node head) {
boolean[] res = new boolean[1];
res[0] = true;
getHeight(head, 1, res);
return res[0];
}
public static int getHeight(Node head, int level, boolean[] res) {
if (head == null) {
return level;
}
int lH = getHeight(head.left, level + 1, res);
if (!res[0]) {
return level;
}
int rH = getHeight(head.right, level + 1, res);
if (!res[0]) {
return level;
}
if (Math.abs(lH - rH) > 1) {
res[0] = false;
}
return Math.max(lH, rH);
}
public static void main(String[] args) {
Node head = new Node(1);
head.left = new Node(2);
head.right = new Node(3);
head.left.left = new Node(4);
head.left.right = new Node(5);
head.right.left = new Node(6);
head.right.right = new Node(7);
System.out.println(isBalance(head));
}
}
判断一棵树是否是搜索二叉树、判断一棵树是否是完全二叉树
- 搜索二叉树（Binary Search Tree）（又：二叉搜索树，二叉排序树）
它或者是一棵空树，或者是具有下列性质的二叉树： 若它的左子树不空，则左子树上所有结点的值均小于它的根结点的值； 若它的右子树不空，则右子树上所有结点的值均大于它的根结点的值； 它的左、右子树也分别为二叉排序树。 - 完全二叉树(Complete Binary Tree)
若设二叉树的深度为h，除第 h 层外，其它各层 (1～h-1) 的结点数都达到最大个数，第 h 层所有的结点都连续集中在最左边，这就是完全二叉树。
/**
* 是否为完全二叉树
*
* @param head
* @return
*/
public static boolean isCBT(Node head) {
if (head == null) {
return true;
}
Queue<Node> queue =new LinkedList<>();
boolean leaf = false;
Node left,right;
//开始时,将头加入队列
queue.offer(head);
while (!queue.isEmpty()) {
//然后在队列中出队当前节点
head = queue.poll();
//拿到左右孩子
left = head.left;
right = head.right;
//开启了叶节点的阶段 且 左/右孩子非空
if ((leaf && (left != null || right != null)) || (left == null && right != null)) {
return false;
}
if (left != null) {
queue.offer(left);
}
if (right != null) {
queue.offer(right);
} else {
leaf = true;
}
}
return true;
}
二叉树层序遍历
