• 算法数据结构体系学习 第十一节

    实现二叉树的按层遍历

    1）其实就是宽度优先遍历，用队列

    2）可以通过设置flag变量的方式，来发现某一层的结束（看题目）

    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;
		}
		Queue<Node> queue = new LinkedList<>();
		queue.add(head);
		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("========");
	}

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    实现二叉树的序列化和反序列化

    1）先序方式序列化和反序列化

    2）按层方式序列化和反序列化

        /*
     * 二叉树可以通过先序、后序或者按层遍历的方式序列化和反序列化，
     * 以下代码全部实现了。
     * 但是，二叉树无法通过中序遍历的方式实现序列化和反序列化
     * 因为不同的两棵树，可能得到同样的中序序列，即便补了空位置也可能一样。
     * 比如如下两棵树
     *         __2
     *        /
     *       1
     *       和
     *       1__
     *          \
     *           2
     * 补足空位置的中序遍历结果都是{ 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;
		}
	}

	public static Queue<String> preSerial(Node head) {
		Queue<String> ans = new LinkedList<>();
		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);
		}
	}

	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);
		}
	}

	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));
		}
	}

	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();
		if (value == null) {
			return null;
		}
		Node head = new Node(Integer.valueOf(value));
		head.left = preb(prelist);
		head.right = preb(prelist);
		return head;
	}

	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;
	}

	public static Queue<String> levelSerial(Node head) {
		Queue<String> ans = new LinkedList<>();
		if (head == null) {
			ans.add(null);
		} else {
			ans.add(String.valueOf(head.value));
			Queue<Node> queue = new LinkedList<Node>();
			queue.add(head);
			while (!queue.isEmpty()) {
				head = queue.poll(); // head 父   子
				if (head.left != null) {
					ans.add(String.valueOf(head.left.value));
					queue.add(head.left);
				} 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;
	}

	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();
			node.left = generateNode(levelList.poll());
			node.right = generateNode(levelList.poll());
			if (node.left != null) {
				queue.add(node.left);
			}
			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!");
		
	}

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    Encode N-ary Tree to Binary Tree

    // 提交时不要提交这个类
	public static class Node {
		public int val;
		public List<Node> children;

		public Node() {
		}

		public Node(int _val) {
			val = _val;
		}

		public Node(int _val, List<Node> _children) {
			val = _val;
			children = _children;
		}
	};

	// 提交时不要提交这个类
	public static class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}

	// 只提交这个类即可
	class Codec {
		// Encodes an n-ary tree to a binary tree.
		public TreeNode encode(Node root) {
			if (root == null) {
				return null;
			}
			TreeNode head = new TreeNode(root.val);
			head.left = en(root.children);
			return head;
		}

		private TreeNode en(List<Node> children) {
			TreeNode head = null;
			TreeNode cur = null;
			for (Node child : children) {
				TreeNode tNode = new TreeNode(child.val);
				if (head == null) {
					head = tNode;
				} else {
					cur.right = tNode;
				}
				cur = tNode;
				cur.left = en(child.children);
			}
			return head;
		}

		// Decodes your binary tree to an n-ary tree.
		public Node decode(TreeNode root) {
			if (root == null) {
				return null;
			}
			return new Node(root.val, de(root.left));
		}

		public List<Node> de(TreeNode root) {
			List<Node> children = new ArrayList<>();
			while (root != null) {
				Node cur = new Node(root.val, de(root.left));
				children.add(cur);
				root = root.right;
			}
			return children;
		}

	}

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    如何设计一个打印整棵树的打印函数

    	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);

	}

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    求二叉树最宽的层有多少个节点

    二叉树结构如下定义：
    Class Node {
    V value;
    Node left;
    Node right;
    Node parent;
    }

    	public static class Node {
		public int value;
		public Node left;
		public Node right;

		public Node(int data) {
			this.value = data;
		}
	}

	public static int maxWidthUseMap(Node head) {
		if (head == null) {
			return 0;
		}
		Queue<Node> queue = new LinkedList<>();
		queue.add(head);
		// key 在 哪一层，value
		HashMap<Node, Integer> levelMap = new HashMap<>();
		levelMap.put(head, 1);
		int curLevel = 1; // 当前你正在统计哪一层的宽度
		int curLevelNodes = 0; // 当前层curLevel层，宽度目前是多少
		int max = 0;
		while (!queue.isEmpty()) {
			Node cur = queue.poll();
			int curNodeLevel = levelMap.get(cur);
			if (cur.left != null) {
				levelMap.put(cur.left, curNodeLevel + 1);
				queue.add(cur.left);
			}
			if (cur.right != null) {
				levelMap.put(cur.right, curNodeLevel + 1);
				queue.add(cur.right);
			}
			if (curNodeLevel == curLevel) {
				curLevelNodes++;
			} else {
				max = Math.max(max, curLevelNodes);
				curLevel++;
				curLevelNodes = 1;
			}
		}
		max = Math.max(max, curLevelNodes);
		return max;
	}

	public static int maxWidthNoMap(Node head) {
		if (head == null) {
			return 0;
		}
		Queue<Node> queue = new LinkedList<>();
		queue.add(head);
		Node curEnd = head; // 当前层，最右节点是谁
		Node nextEnd = null; // 下一层，最右节点是谁
		int max = 0;
		int curLevelNodes = 0; // 当前层的节点数
		while (!queue.isEmpty()) {
			Node cur = queue.poll();
			if (cur.left != null) {
				queue.add(cur.left);
				nextEnd = cur.left;
			}
			if (cur.right != null) {
				queue.add(cur.right);
				nextEnd = cur.right;
			}
			curLevelNodes++;
			if (cur == curEnd) {
				max = Math.max(max, curLevelNodes);
				curLevelNodes = 0;
				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!");

	}

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    给你二叉树中的某个节点，返回该节点的后继节点

    限定中序遍历

    public static class Node {
		public int value;
		public Node left;
		public Node right;
		public Node parent;

		public Node(int data) {
			this.value = data;
		}
	}

	public static Node getSuccessorNode(Node node) {
		if (node == null) {
			return node;
		}
		if (node.right != null) {
			return getLeftMost(node.right);
		} else { // 无右子树
			Node parent = node.parent;
			while (parent != null && parent.right == node) { // 当前节点是其父亲节点右孩子
				node = parent;
				parent = node.parent;
			}
			return parent;
		}
	}

	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));
	}

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    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

    	public static void printAllFolds(int N) {
		process(1, N, true);
		System.out.println();
	}

	// 当前你来了一个节点，脑海中想象的！
	// 这个节点在第i层，一共有N层，N固定不变的
	// 这个节点如果是凹的话，down = T
	// 这个节点如果是凸的话，down = F
	// 函数的功能：中序打印以你想象的节点为头的整棵树！
	public static void process(int i, int N, boolean down) {
		if (i > N) {
			return;
		}
		process(i + 1, N, true);
		System.out.print(down ? "凹 " : "凸 ");
		process(i + 1, N, false);
	}

	public static void main(String[] args) {
		int N = 4;
		printAllFolds(N);
	}

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  • 原文地址：https://blog.csdn.net/qq_29374433/article/details/126325451