第一题
实现二叉树的先序、中序、后序遍历,包括递归方式和非递归方式
第二题
如何直观的打印一颗二叉树
解析
为了能够更加直观的看见一颗二叉树而设计的程序
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);
}
}
第三题
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;
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。
题目四
介绍二叉树的序列化和反序列化
题目五
折纸问题
请把一段纸条竖着放在桌子上,然后从纸条的下边向上方对折1次,压出折痕后展开。此时 折痕是凹下去的,即折痕突起的方向指向纸条的背面。如果从纸条的下边向上方连续对折2 次,压出折痕后展开,此时有三条折痕,从上到下依次是下折痕、下折痕和上折痕。给定一 个输入参数N,代表纸条都从下边向上方连续对折N次,请从上到下打印所有折痕的方向。
例如:N=1时,打印: down
N=2时,打印: down down up
N=3时,打印: down down up down down up up
N=4时,打印: down down up down down up up down down down up up down up up
解答
当遍历到右边子树时,需要标记打印出up,这和中序遍历类似。如果不理解可以使用前序遍历和后续遍历。
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);
}
}
题目六
判断一棵二叉树是否是平衡二叉树
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
}
}
题目七
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
解析
利用了二叉树的前序遍历的顺序和中序遍历的顺序,在前序遍历时,获得右子树的值,中序遍历获得左子树的值,然后在后续遍历中二者比较。
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;
}
}
题目八
已知一棵完全二叉树,求其节点的个数
要求:时间复杂度低于O(N),N为这棵树的节点个数