左神第四課
第一題
實現二叉樹的先序、中序、後序遍歷,包括遞迴方式和非遞迴方式
第二題
如何直觀的列印一顆二叉樹
解析
為了能夠更加直觀的看見一顆二叉樹而設計的程式
public class Code_02_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);
}
}
第三題
在二叉樹中找到一個節點的後繼節點
現在有一種新的二叉樹節點型別如下:
public class Node {
public int value;
public Node left;
public Node right;
public Node parent;
public Node(int data) { this.value = data; }
}
該結構比普通二叉樹節點結構多了一個指向父節點的parent指標。假設有一 棵Node型別的節點組成的二叉樹,樹中每個節點的parent指標都正確地指向 自己的父節點,頭節點的parent指向null。只給一個在二叉樹中的某個節點 node,請實現返回node的後繼節點的函式。在二叉樹的中序遍歷的序列中, node的下一個節點叫作node的後繼節點。
題目四
介紹二叉樹的序列化和反序列化
題目五
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,即摺痕突起的方向指向紙條的背面。如果從紙條的下邊向上方連續對摺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摺痕的方向。
例如:N=1時,列印: down
N=2時,列印: down down up
N=3時,列印: down down up down down up up
N=4時,列印: down down up down down up up down down down up up down up up
解答
當遍歷到右邊子樹時,需要標記打印出up,這和中序遍歷類似。如果不理解可以使用前序遍歷和後續遍歷。
public class Code_05_PaperFolding {
public static void printAllFolds(int N) {
printProcess(N, true);
}
public static void printProcess(int N, boolean down) {
if (N<=0) {
return;
}
printProcess(N-1, true);
System.out.println(down ? "down " : "up ");
printProcess(N-1, false);
}
public static void main(String[] args) {
int N = 3;
printAllFolds(N);
}
}
題目六
判斷一棵二叉樹是否是平衡二叉樹
解析
分治法
public class Test2 {
class ReturnType{
private int high;
private boolean isBalance;
public ReturnType() {
super();
}
public ReturnType(int high, boolean isBalance) {
this.high = high;
this.isBalance = isBalance;
}
public int getHigh() {
return high;
}
public void setHigh(int high) {
this.high = high;
}
public boolean isBalance() {
return isBalance;
}
public void setBalance(boolean isBalance) {
this.isBalance = isBalance;
}
}
public boolean isBalanced(TreeNode root) {
return isBlanced2(root).isBalance();
}
public ReturnType isBlanced2(TreeNode root){
if (root==null) {
return new ReturnType(0, true);
}
ReturnType left=isBlanced2(root.left);
ReturnType right=isBlanced2(root.right);
ReturnType r=new ReturnType();
if (left.isBalance&&right.isBalance) {
r.setBalance(Math.abs(left.high-right.high)<=1);
r.setHigh(Math.max(left.high, right.high)+1);
return r;
}
return new ReturnType(-1, false);
}
public static void main(String[] args) {
// TODO Auto-generated method stub
}
}
題目七
判斷一棵樹是否是搜尋二叉樹、判斷一棵樹是否是完全二叉樹
解析
利用了二叉樹的前序遍歷的順序和中序遍歷的順序,在前序遍歷時,獲得右子樹的值,中序遍歷獲得左子樹的值,然後在後續遍歷中二者比較。
分治法也是可以的。
class Solution {
public boolean isValidBST(TreeNode root) {
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
private boolean isValidBST(TreeNode root, long minVal, long maxVal)
{
if (root == null) {
return true;
}
int c=root.val;
boolean a = isValidBST(root.left, minVal, c);
int d=root.val;
boolean b = isValidBST(root.right, d, maxVal);
if (root.val >= maxVal || root.val <= minVal) {
return false;
}
return a && b;
}
}
題目八
已知一棵完全二叉樹,求其節點的個數
要求:時間複雜度低於O(N),N為這棵樹的節點個數