《演算法筆記》7. 二叉樹基本演算法整理
目錄
1 二叉樹基本演算法
1.1 二叉樹的遍歷
1.1.1 二叉樹節點定義
Class Node{ // 節點的值型別 V value; // 二叉樹的左孩子指標 Node left; // 二叉樹的右孩子指標 Node right; }
1.1.2 遞迴實現先序中序後序遍歷
先序:任何子樹的處理順序都是,先頭結點,再左子樹,再右子樹。先處理頭結點
中序:任何子樹的處理順序都是,先左子樹,再頭結點,再右子樹。中間處理頭結點
後序:任何子樹的處理順序都是,先左子樹,再右子樹,再頭結點。最後處理頭結點
對於下面的一棵樹:
graph TD
1-->2
1-->3
2-->4
2-->5
3-->6
3-->7
1、 先序遍歷為:1 2 4 5 3 6 7
2、 中序遍歷為:4 2 5 1 6 3 7
3、 後序遍歷為:4 5 2 6 7 3 1
package class07; public class Code01_RecursiveTraversalBT { public static class Node { public int value; public Node left; public Node right; public Node(int v) { value = v; } } public static void f(Node head) { if (head == null) { return; } // 1 此處列印等於先序 f(head.left); // 2 此處列印等於中序 f(head.right); // 3 此處列印等於後序 } // 先序列印所有節點 public static void pre(Node head) { if (head == null) { return; } // 列印頭 System.out.println(head.value); // 遞迴列印左子樹 pre(head.left); // 遞迴列印右子樹 pre(head.right); } // 中序遍歷 public static void in(Node head) { if (head == null) { return; } in(head.left); System.out.println(head.value); in(head.right); } // 後序遍歷 public static void pos(Node head) { if (head == null) { return; } pos(head.left); pos(head.right); System.out.println(head.value); } 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); pre(head); System.out.println("========"); in(head); System.out.println("========"); pos(head); System.out.println("========"); } }
結論:對於樹的遞迴,每個節點實質上會到達三次,例如上文的樹結構,對於f函式,我們傳入頭結點,再呼叫左樹再呼叫右樹。實質上經過的路徑為1 2 4 4 4 2 5 5 5 2 1 3 6 6 6 3 7 7 7 3 1。我們在每個節點三次返回的基礎上,第一次到達該節點就列印,就是先序,第二次到達該節點列印就是中序,第三次到達該節點就是後序。
所以先序中序後序,只是我們的遞迴順序加工出來的結果!
1.1.3 非遞迴實現先序中序後序遍歷
思路:由於任何遞迴可以改為非遞迴,我們可以使用壓棧來實現。用先序實現的步驟,其他類似:
步驟一,把節點壓入棧中,彈出就列印
步驟二,如果有右孩子先壓入右孩子
步驟三,如果有左孩子壓入左孩子
package class07;
import java.util.Stack;
public class Code02_UnRecursiveTraversalBT {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int v) {
value = v;
}
}
// 非遞迴先序
public static void pre(Node head) {
System.out.print("pre-order: ");
if (head != null) {
Stack<Node> stack = new Stack<Node>();
stack.add(head);
while (!stack.isEmpty()) {
// 彈出就列印
head = stack.pop();
System.out.print(head.value + " ");
// 右孩子不為空,壓右
if (head.right != null) {
stack.push(head.right);
}
// 左孩子不為空,壓左
if (head.left != null) {
stack.push(head.left);
}
}
}
System.out.println();
}
// 非遞迴中序
public static void in(Node head) {
System.out.print("in-order: ");
if (head != null) {
Stack<Node> stack = new Stack<Node>();
while (!stack.isEmpty() || head != null) {
// 整條左邊界依次入棧
if (head != null) {
stack.push(head);
head = head.left;
// 左邊界到頭彈出一個列印,來到該節點右節點,再把該節點的左樹以此進棧
} else {
head = stack.pop();
System.out.print(head.value + " ");
head = head.right;
}
}
}
System.out.println();
}
// 非遞迴後序
public static void pos1(Node head) {
System.out.print("pos-order: ");
if (head != null) {
Stack<Node> s1 = new Stack<Node>();
// 輔助棧
Stack<Node> s2 = new Stack<Node>();
s1.push(head);
while (!s1.isEmpty()) {
head = s1.pop();
s2.push(head);
if (head.left != null) {
s1.push(head.left);
}
if (head.right != null) {
s1.push(head.right);
}
}
while (!s2.isEmpty()) {
System.out.print(s2.pop().value + " ");
}
}
System.out.println();
}
// 非遞迴後序2:用一個棧實現後序遍歷,比較有技巧
public static void pos2(Node h) {
System.out.print("pos-order: ");
if (h != null) {
Stack<Node> stack = new Stack<Node>();
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();
}
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);
pre(head);
System.out.println("========");
in(head);
System.out.println("========");
pos1(head);
System.out.println("========");
pos2(head);
System.out.println("========");
}
}
1.1.4 二叉樹按層遍歷
1、 其實就是寬度優先遍歷,用佇列
2、 可以通過設定flag變數的方式,來發現某一層的結束
按層列印輸出二叉樹
package class07;
import java.util.LinkedList;
import java.util.Queue;
public class Code03_LevelTraversalBT {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int v) {
value = v;
}
}
public static void level(Node head) {
if (head == null) {
return;
}
// 準備一個輔助佇列
Queue<Node> queue = new LinkedList<>();
// 加入頭結點
queue.add(head);
// 佇列不為空出隊列印,把當前節點的左右孩子加入佇列
while (!queue.isEmpty()) {
Node cur = queue.poll();
System.out.println(cur.value);
if (cur.left != null) {
queue.add(cur.left);
}
if (cur.right != null) {
queue.add(cur.right);
}
}
}
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);
level(head);
System.out.println("========");
}
}
找到二叉樹的最大寬度
package class07;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Queue;
public class Code06_TreeMaxWidth {
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
// 方法1使用map
public static int maxWidthUseMap(Node head) {
if (head == null) {
return 0;
}
Queue<Node> queue = new LinkedList<>();
queue.add(head);
// key(節點) 在 哪一層,value
HashMap<Node, Integer> levelMap = new HashMap<>();
// head在第一層
levelMap.put(head, 1);
// 當前你正在統計哪一層的寬度
int curLevel = 1;
// 當前層curLevel層,寬度目前是多少
int curLevelNodes = 0;
// 用來儲存所有層的最大值,也就是最大寬度
int max = 0;
while (!queue.isEmpty()) {
Node cur = queue.poll();
int curNodeLevel = levelMap.get(cur);
// 當前節點的左孩子不為空,佇列加入左孩子,層數在之前層上加1
if (cur.left != null) {
levelMap.put(cur.left, curNodeLevel + 1);
queue.add(cur.left);
}
// 當前節點的右孩子不為空,佇列加入右孩子,層數也變為當前節點的層數加1
if (cur.right != null) {
levelMap.put(cur.right, curNodeLevel + 1);
queue.add(cur.right);
}
// 當前層等於正在統計的層數,不結算
if (curNodeLevel == curLevel) {
curLevelNodes++;
} else {
// 新的一層,需要結算
// 得到目前為止的最大寬度
max = Math.max(max, curLevelNodes);
curLevel++;
// 結算後,當前層節點數設定為1
curLevelNodes = 1;
}
}
// 由於最後一層,沒有新的一層去結算,所以這裡單獨結算最後一層
max = Math.max(max, curLevelNodes);
return max;
}
// 方法2不使用map
public static int maxWidthNoMap(Node head) {
if (head == null) {
return 0;
}
Queue<Node> queue = new LinkedList<>();
queue.add(head);
// 當前層,最右節點是誰,初始head的就是本身
Node curEnd = head;
// 如果有下一層,下一層最右節點是誰
Node nextEnd = null;
// 全域性最大寬度
int max = 0;
// 當前層的節點數
int curLevelNodes = 0;
while (!queue.isEmpty()) {
Node cur = queue.poll();
// 左邊不等於空,加入左
if (cur.left != null) {
queue.add(cur.left);
// 孩子的最右節點暫時為左節點
nextEnd = cur.left;
}
// 右邊不等於空,加入右
if (cur.right != null) {
queue.add(cur.right);
// 如果有右節點,孩子層的最右要更新為右節點
nextEnd = cur.right;
}
// 由於最開始彈出當前節點,那麼該層的節點數加一
curLevelNodes++;
// 當前節點是當前層最右的節點,進行結算
if (cur == curEnd) {
// 當前層的節點和max進行比較,計算當前最大的max
max = Math.max(max, curLevelNodes);
// 即將進入下一層,重置下一層節點為0個節點
curLevelNodes = 0;
// 當前層的最右,直接更新為找出來的下一層最右
curEnd = nextEnd;
}
}
return max;
}
// for test
public static Node generateRandomBST(int maxLevel, int maxValue) {
return generate(1, maxLevel, maxValue);
}
// for test
public static Node generate(int level, int maxLevel, int maxValue) {
if (level > maxLevel || Math.random() < 0.5) {
return null;
}
Node head = new Node((int) (Math.random() * maxValue));
head.left = generate(level + 1, maxLevel, maxValue);
head.right = generate(level + 1, maxLevel, maxValue);
return head;
}
public static void main(String[] args) {
int maxLevel = 10;
int maxValue = 100;
int testTimes = 1000000;
for (int i = 0; i < testTimes; i++) {
Node head = generateRandomBST(maxLevel, maxValue);
if (maxWidthUseMap(head) != maxWidthNoMap(head)) {
System.out.println("Oops!");
}
}
System.out.println("finish!");
}
}
1.2 二叉樹的序列化和反序列化
1、 可以用先序或者中序或者後序或者按層遍歷,來實現二叉樹的序列化
2、 用了什麼方式的序列化,就用什麼方式的反序列化
由於如果樹上的節點值相同,那麼序列化看不出來該樹的結構,所以我們的序列化要加上空間結構的標識,空節點補全的方式。
package class07;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
public class Code04_SerializeAndReconstructTree {
/*
* 二叉樹可以通過先序、後序或者按層遍歷的方式序列化和反序列化,
* 以下程式碼全部實現了。
* 但是,二叉樹無法通過中序遍歷的方式實現序列化和反序列化
* 因為不同的兩棵樹,可能得到同樣的中序序列,即便補了空位置也可能一樣。
* 比如如下兩棵樹
* __2
* /
* 1
* 和
* 1__
* \
* 2
* 補足空位置的中序遍歷結果都是{ null, 1, null, 2, null}
*
* */
public static class Node {
public int value;
public Node left;
public Node right;
public Node(int data) {
this.value = data;
}
}
// 先序序列化
public static Queue<String> preSerial(Node head) {
Queue<String> ans = new LinkedList<>();
// 先序的序列化結果依次放入佇列中去
pres(head, ans);
return ans;
}
public static void pres(Node head, Queue<String> ans) {
if (head == null) {
ans.add(null);
} else {
ans.add(String.valueOf(head.value));
pres(head.left, ans);
pres(head.right, ans);
}
}
// 中序有問題。見檔案開頭註釋
public static Queue<String> inSerial(Node head) {
Queue<String> ans = new LinkedList<>();
ins(head, ans);
return ans;
}
public static void ins(Node head, Queue<String> ans) {
if (head == null) {
ans.add(null);
} else {
ins(head.left, ans);
ans.add(String.valueOf(head.value));
ins(head.right, ans);
}
}
// 後序序列化
public static Queue<String> posSerial(Node head) {
Queue<String> ans = new LinkedList<>();
poss(head, ans);
return ans;
}
public static void poss(Node head, Queue<String> ans) {
if (head == null) {
ans.add(null);
} else {
poss(head.left, ans);
poss(head.right, ans);
ans.add(String.valueOf(head.value));
}
}
// 根據先序的結構,構建這顆樹
public static Node buildByPreQueue(Queue<String> prelist) {
if (prelist == null || prelist.size() == 0) {
return null;
}
return preb(prelist);
}
public static Node preb(Queue<String> prelist) {
String value = prelist.poll();
// 如果頭節點是空的話,返回空
if (value == null) {
return null;
}
// 否則根據第一個值構建先序的頭結點
Node head = new Node(Integer.valueOf(value));
// 遞迴建立左樹
head.left = preb(prelist);
// 遞迴建立右樹
head.right = preb(prelist);
return head;
}
// 根據後序的結構,構建該樹
public static Node buildByPosQueue(Queue<String> poslist) {
if (poslist == null || poslist.size() == 0) {
return null;
}
// 左右中 -> stack(中右左)
Stack<String> stack = new Stack<>();
while (!poslist.isEmpty()) {
stack.push(poslist.poll());
}
return posb(stack);
}
public static Node posb(Stack<String> posstack) {
String value = posstack.pop();
if (value == null) {
return null;
}
Node head = new Node(Integer.valueOf(value));
head.right = posb(posstack);
head.left = posb(posstack);
return head;
}
// 按層序列化,整體上就是寬度優先遍歷
public static Queue<String> levelSerial(Node head) {
// 序列化結果
Queue<String> ans = new LinkedList<>();
if (head == null) {
ans.add(null);
} else {
// 加入一個節點的時候,把該節點的值加入
ans.add(String.valueOf(head.value));
// 輔助佇列
Queue<Node> queue = new LinkedList<Node>();
queue.add(head);
while (!queue.isEmpty()) {
head = queue.poll();
// 左孩子不為空,即序列化,也加入佇列
if (head.left != null) {
ans.add(String.valueOf(head.left.value));
queue.add(head.left);
// 左孩子等於空,只序列化,不加入佇列
} else {
ans.add(null);
}
if (head.right != null) {
ans.add(String.valueOf(head.right.value));
queue.add(head.right);
} else {
ans.add(null);
}
}
}
return ans;
}
// 按層反序列化
public static Node buildByLevelQueue(Queue<String> levelList) {
if (levelList == null || levelList.size() == 0) {
return null;
}
Node head = generateNode(levelList.poll());
Queue<Node> queue = new LinkedList<Node>();
if (head != null) {
queue.add(head);
}
Node node = null;
while (!queue.isEmpty()) {
node = queue.poll();
// 不管左右孩子是否為空,都要加節點
node.left = generateNode(levelList.poll());
node.right = generateNode(levelList.poll());
// 左孩子不為空,佇列加左,為建下一層做準備
if (node.left != null) {
queue.add(node.left);
}
// 右孩子不為空,佇列加右,為建下一層做準備
if (node.right != null) {
queue.add(node.right);
}
}
return head;
}
public static Node generateNode(String val) {
if (val == null) {
return null;
}
return new Node(Integer.valueOf(val));
}
// for test
public static Node generateRandomBST(int maxLevel, int maxValue) {
return generate(1, maxLevel, maxValue);
}
// for test
public static Node generate(int level, int maxLevel, int maxValue) {
if (level > maxLevel || Math.random() < 0.5) {
return null;
}
Node head = new Node((int) (Math.random() * maxValue));
head.left = generate(level + 1, maxLevel, maxValue);
head.right = generate(level + 1, maxLevel, maxValue);
return head;
}
// for test
public static boolean isSameValueStructure(Node head1, Node head2) {
if (head1 == null && head2 != null) {
return false;
}
if (head1 != null && head2 == null) {
return false;
}
if (head1 == null && head2 == null) {
return true;
}
if (head1.value != head2.value) {
return false;
}
return isSameValueStructure(head1.left, head2.left) && isSameValueStructure(head1.right, head2.right);
}
// for test
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) {
int maxLevel = 5;
int maxValue = 100;
int testTimes = 1000000;
System.out.println("test begin");
for (int i = 0; i < testTimes; i++) {
Node head = generateRandomBST(maxLevel, maxValue);
Queue<String> pre = preSerial(head);
Queue<String> pos = posSerial(head);
Queue<String> level = levelSerial(head);
Node preBuild = buildByPreQueue(pre);
Node posBuild = buildByPosQueue(pos);
Node levelBuild = buildByLevelQueue(level);
if (!isSameValueStructure(preBuild, posBuild) || !isSameValueStructure(posBuild, levelBuild)) {
System.out.println("Oops!");
}
}
System.out.println("test finish!");
}
}
1.3 直觀列印一顆二叉樹
如何設計一個列印整顆數的列印函式,簡單起見,我們躺著列印,正常的樹我們順時針旋轉90°即可
package class07;
public class Code05_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:");
// 列印函式,先傳入頭結點
printInOrder(head, 0, "H", 17);
System.out.println();
}
// head表示當前傳入節點
// height當前節點所在的高度
// to表示當前節點的指向資訊
// len表示列印當前值填充到多少位當成一個完整的值
public static void printInOrder(Node head, int height, String to, int len) {
if (head == null) {
return;
}
// 遞迴右樹,右樹向下指
printInOrder(head.right, height + 1, "v", len);
/**
* 列印自己的值
* val 表示值內容
**/
String val = to + head.value + to;
int lenM = val.length();
// 按照len算該值左側需要填充多少空格
int lenL = (len - lenM) / 2;
// 按照len算該值右側需要填充多少空格
int lenR = len - lenM - lenL;
// 實際值加上左右佔位,表示每個值包括佔位之後大小
val = getSpace(lenL) + val + getSpace(lenR);
System.out.println(getSpace(height * len) + val);
// 遞迴左樹,左樹向上指
printInOrder(head.left, height + 1, "^", len);
}
// 根據height*len補空格
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);
}
}
1.4 題目實戰
1.4.1 題目一:返回二叉樹的後繼節點
題目描述:二叉樹的結構定義如下:
Class Node {
V value;
Node left;
Node right;
// 指向父親節點
Node parent;
}
給你二叉樹中的某個節點,返回該節點的後繼節點。後繼節點表示一顆二叉樹中,在中序遍歷的序列中,一個個節點的下一個節點是誰。
方法一,通常解法思路:由於我們的節點有指向父節點的指標,而整顆二叉樹的頭結點的父節點為null。那麼我們可以找到整棵樹的頭結點,然後中序遍歷,再找到給定節點的下一個節點,就是該節點的後續節點。
方法二,考慮一個節點和其後繼節點的結構之間的關係:
如果一個節點x有右樹,那麼其後繼節點就是右樹最左的節點。
如果x沒有右樹,往上找父親節點。如果x是其父親的右孩子繼續往上找,如果某節點是其父親節點的左孩子,那麼該節點的父親就是x的後繼節點
即如果某節點左樹的最右節點是x,那麼該節點是x的後繼
如果找父節點,一直找到null都不滿足,那麼該節點是整棵樹的最右節點,沒有後繼
package class07;
public class Code07_SuccessorNode {
public static class Node {
public int value;
public Node left;
public Node right;
public Node parent;
public Node(int data) {
this.value = data;
}
}
// 給定節點,返回後繼
public static Node getSuccessorNode(Node node) {
if (node == null) {
return node;
}
if (node.right != null) {
return getLeftMost(node.right);
// 無右子樹
} else {
Node parent = node.parent;
// 當前節點是其父親節點右孩子,繼續
while (parent != null && parent.right == node) {
node = parent;
parent = node.parent;
}
return parent;
}
}
// 找右樹上的最左節點
public static Node getLeftMost(Node node) {
if (node == null) {
return node;
}
while (node.left != null) {
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));
}
}
後繼節點對應的是前驅結點,前驅結點的含義是中序遍歷,某節點的前一個節點
1.4.2 題目二:摺紙問題
請把一段紙條豎著放在桌子上,然後從紙條的下邊向上方對摺1次,壓出摺痕後展開。
此時摺痕是凹下去的,即摺痕凸起的方向指向紙條的背面。
如果從紙條的下邊向上方對摺2次,壓出摺痕後展開,此時有三條摺痕,從上到下依次是下摺痕,下摺痕和上摺痕。
給定一個輸入引數N,代表紙條都從下邊向上方連續對摺N次。請從上到下列印所有的摺痕的方向。
例如:N=1時,列印: down 。N=2時,列印:down down up
規律,大於一次後,每次摺痕出現的位置都是在上次摺痕的上方出現凹摺痕,下方出現凸摺痕。所以我們沒必要構建這顆樹,就可以用遞迴思維解決
對應的樹結構按層輸出為:
1凹
2凹 2凸
3凹 3凸 3凹 3凸
package class07;
public class Code08_PaperFolding {
public static void printAllFolds(int N) {
// 先從頭結點出發,i初始值為1,切第一次的頭結點摺痕為凹摺痕
printProcess(1, N, true);
}
// 遞迴過程,來到了某一個節點,
// i是節點的層數,N一共的層數,down == true 凹 down == false 凸
public static void printProcess(int i, int N, boolean down) {
if (i > N) {
return;
}
// 每個當前節點的左子節點是凹
printProcess(i + 1, N, true);
System.out.println(down ? "凹 " : "凸 ");
// 每個當前節點的右子樹是凸
printProcess(i + 1, N, false);
}
public static void main(String[] args) {
int N = 3;
// 折N次,列印所有凹凸分佈情況
printAllFolds(N);
}
}