Java与算法（9）

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Java与算法（9）

1.给定二叉树头节点，每个节点的值都不同，找到含有节点最多的搜索二叉子树。

public class FindBST {
	public static class Node {
		int data;
		Node left;
		Node right;
		public Node(int data) {
			// TODO Auto-generated constructor stub
			this.data = data;
			left = null;
			right = null;
		}
	}
	
	public static ArrayList<Node> trees = new ArrayList<>();
	public void buildTree(int[] array) {
		for(int i=0;i<array.length;i++) {
		trees.add(new Node(array[i]));
		}
		
		for(int index = 0;index<array.length/2-1;index++) {
			trees.get(index).left =  trees.get(index*2+1);
			trees.get(index).right = trees.get(index*2+2);
		}
		int lastindex = array.length/2-1;
		trees.get(lastindex).left = trees.get(lastindex*2+1);
		if (array.length%2==1) {
			trees.get(lastindex).right = trees.get(lastindex*2+2);
			
		}
		
	}
	
	public Node bigBST(Node root) {
		int[] record =  new int[3];
		return find(root, record);
	}
	
	public Node find(Node root,int[] r) {
		if (root==null) {
			r[0] = 0;
			r[1] = Integer.MAX_VALUE;
			r[2] = Integer.MAX_VALUE;
			return null;
		}
		
		int data = root.data;
		Node right = root.right;
		Node left = root.left;
		Node lBST = find(left,r);
		int lSIZE = r[0];
		int lMIN = r[1];
		int lMAX = r[2];
		Node rBST = find(right, r);
		int rSIZE = r[0];
		int rMIN = r[1];
		int rMAX = r[2];
		r[1] = Math.min(lMIN, data);
		r[2] = Math.max(rMAX, data);
		if (left==lBST && right==rBST && lMAX<data && data<rMIN) {
			r[0] = rSIZE+lSIZE +1;
			return root;
		}
		r[0] = Math.max(rSIZE, lSIZE);
		return lSIZE>rSIZE?lBST:rBST;
	}
	

}

2.实现二叉树三种遍历，要求：

时间复杂度O(N)，空间复杂度O(1)

public class MorrisTree {
	public static class Node {
		int data;
		Node left;
		Node right;
		public Node(int data) {
			// TODO Auto-generated constructor stub
			this.data = data;
			left = null;
			right = null;
		}
	}
	
	public static ArrayList<Node> trees = new ArrayList<>();
	public void buildTree(int[] array) {
		for(int i=0;i<array.length;i++) {
		trees.add(new Node(array[i]));
		}
		
		for(int index = 0;index<array.length/2-1;index++) {
			trees.get(index).left =  trees.get(index*2+1);
			trees.get(index).right = trees.get(index*2+2);
		}
		int lastindex = array.length/2-1;
		trees.get(lastindex).left = trees.get(lastindex*2+1);
		if (array.length%2==1) {
			trees.get(lastindex).right = trees.get(lastindex*2+2);
			
		}
		
	}
	
	
	/**
	 * 先序遍历
	 * @param head
	 */
	public static void morrisIn(Node head) {
		if (head == null) {
			return;
		}
		Node cur1 = head;
		Node cur2 = null;
		while (cur1 != null) {
			cur2 = cur1.left;
			if (cur2 != null) {
				while (cur2.right != null && cur2.right != cur1) {
					cur2 = cur2.right;
				}
				if (cur2.right == null) {
					cur2.right = cur1;
					cur1 = cur1.left;
					continue;
				} else {
					cur2.right = null;
				}
			}
			System.out.print(cur1.data + " ");
			cur1 = cur1.right;
		}
		System.out.println();
	}
	
	
	/**
	 * 中序遍历
	 * @param head
	 */
	public static void morrisPre(Node head) {
		if (head == null) {
			return;
		}
		Node cur1 = head;
		Node cur2 = null;
		while (cur1 != null) {
			cur2 = cur1.left;
			if (cur2 != null) {
				while (cur2.right != null && cur2.right != cur1) {
					cur2 = cur2.right;
				}
				if (cur2.right == null) {
					cur2.right = cur1;
					System.out.print(cur1.data + " ");
					cur1 = cur1.left;
					continue;
				} else {
					cur2.right = null;
				}
			} else {
				System.out.print(cur1.data + " ");
			}
			cur1 = cur1.right;
		}
		System.out.println();
	}
	
	/**
	 * 后序遍历
	 * @param head
	 */
	public static void morrisPos(Node head) {
		if (head == null) {
			return;
		}
		Node cur1 = head;
		Node cur2 = null;
		while (cur1 != null) {
			cur2 = cur1.left;
			if (cur2 != null) {
				while (cur2.right != null && cur2.right != cur1) {
					cur2 = cur2.right;
				}
				if (cur2.right == null) {
					cur2.right = cur1;
					cur1 = cur1.left;
					continue;
				} else {
					cur2.right = null;
					printEdge(cur1.left);
				}
			}
			cur1 = cur1.right;
		}
		printEdge(head);
		System.out.println();
	}

	public static void printEdge(Node head) {
		Node tail = reverseEdge(head);
		Node cur = tail;
		while (cur != null) {
			System.out.print(cur.data + " ");
			cur = cur.right;
		}
		reverseEdge(tail);
	}

	public static Node reverseEdge(Node from) {
		Node pre = null;
		Node next = null;
		while (from != null) {
			next = from.right;
			from.right = pre;
			pre = from;
			from = next;
		}
		return pre;
	}
	
	
	

	public static void main(String[] args) {
		int[] array = {3,6,5,9,10,7,29};
		MorrisTree tree = new MorrisTree();
		tree.buildTree(array);
		tree.morrisIn(trees.get(0));
		System.out.println();
		tree.morrisPre(trees.get(0));
		System.out.println();
		tree.morrisPos(trees.get(0));
	}

}

3.实现二叉树三种遍历（递归和非递归）

public class BinartTree {
	
	public static class Node {
		int data;
		Node left;
		Node right;
		public Node(int data) {
			// TODO Auto-generated constructor stub
			this.data = data;
			left = null;
			right = null;
		}
	}
	public static ArrayList<Node> trees = new ArrayList<>();
	public void buildTree(int[] array) {
		for(int i=0;i<array.length;i++) {
		trees.add(new Node(array[i]));
		}
		
		for(int index = 0;index<array.length/2-1;index++) {
			trees.get(index).left =  trees.get(index*2+1);
			trees.get(index).right = trees.get(index*2+2);
		}
		int lastindex = array.length/2-1;
		trees.get(lastindex).left = trees.get(lastindex*2+1);
		if (array.length%2==1) {
			trees.get(lastindex).right = trees.get(lastindex*2+2);
			
		}
		
		
	}
	
	/**
	 * 先序遍历,递归实现
	 * @param root
	 */
	public void preOrder(Node root) {
		if (root==null) {
			return;
		}
		System.out.print(root.data+" ");
		preOrder(root.left);
		preOrder(root.right);
	}
	
	/**
	 * 先序遍历，非递归实现
	 * @param root
	 */
	public void preOrderNotrecur(Node root) {
		Stack<Node> stack = new Stack<>();
		if (root!=null) {
			stack.push(root);
			while (!stack.isEmpty()) {
				root = stack.pop();
				System.out.print(root.data+" ");
				if (root.right!=null) {
					stack.push(root.right);
				}
				if (root.left!=null) {
					stack.push(root.left);
				}
			}
			
		}
		System.out.println("");
		
		
	}
	
	
	/**
	 * 中序遍历，递归实现
	 * @param root
	 */
	public void inOrder(Node root) {
		if (root == null) {
			return;
		}
		inOrder(root.left);
		System.out.print(root.data+" ");
		inOrder(root.right);
	}
	
	/**
	 * 中序遍历，非递归实现
	 * @param root
	 */
	public void inOrderNotrecur(Node root) {
		Stack<Node> stack = new Stack<>();
		if (root!=null) {
			while (!stack.isEmpty() || root!=null) {
				if (root!=null) {
					stack.push(root);
					root=root.left;
				} else {
					root = stack.pop();
					System.out.print(root.data+" ");
					root = root.right;
				}
				
			}
			
		}
		System.out.println();
		
	}
	
	
	/**
	 * 后序遍历，递归实现
	 * @param root
	 */
	public void posOrder(Node root) {
		if (root == null) {
			return ;
		}
		
		posOrder(root.left);
		posOrder(root.right);
		System.out.print(root.data+" ");
	}
	
	public void posOrderNotrecur(Node root) {
		Stack<Node> stack = new Stack<>();
		if (root!=null) {
			stack.push(root);
			Node node = null;
			while (!stack.isEmpty()) {
				node=stack.peek();
				if (node.left!=null && root!=node.left && root!=node.right) {
						stack.push(node.left);
				} else if (node.right!=null && root!=node.right) {
					stack.push(node.right);
				} else {
					System.out.print(stack.pop().data+" ");
					root = node;
				}
				
				
			}
			
			
		}
		
		System.out.println();
		
	}
	
	
	public static void main(String[] args) {
			int[] array = {3,6,5,9,10,7,29};
			BinartTree tree = new BinartTree();
			tree.buildTree(array);
			System.out.println("先序遍历递归实现");
			tree.preOrder(trees.get(0));
			System.out.println("");
			System.out.println("先序遍历非递归实现");
			tree.preOrderNotrecur(trees.get(0));
			System.out.println();
			System.out.println("中序遍历递归实现");
			tree.inOrder(trees.get(0));
			System.out.println();
			System.out.println("中序遍历递归实现");
			tree.inOrderNotrecur(trees.get(0));
			System.out.println();
			System.out.println("后序遍历递归实现");
			tree.posOrder(trees.get(0));
			System.out.println("");
			System.out.println("后序遍历递归实现");
			tree.posOrderNotrecur(trees.get(0));
	}

}