数据结构（二叉树的递归套路**）（8）

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前言

数据结构和算法笔记

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一、给定一棵二叉树的头节点head，返回这颗二叉树是不是平衡二叉树

public static boolean isBalanced2(Node head) {
		return process(head).isBalanced;
	}
	
	public static class Info{
		public boolean isBalanced;
		public int height;
		
		public Info(boolean i, int h) {
			isBalanced = i;
			height = h;
		}
	}
	
	public static Info process(Node x) {
		if(x == null) {
			return new Info(true, 0);
		}
		Info leftInfo = process(x.left);
		Info rightInfo = process(x.right);
		int height = Math.max(leftInfo.height, rightInfo.height)  + 1;
		boolean isBalanced = true;
		if(!leftInfo.isBalanced) {
			isBalanced = false;
		}
		if(!rightInfo.isBalanced) {
			isBalanced = false;
		}
		if(Math.abs(leftInfo.height - rightInfo.height) > 1) {
			isBalanced = false;
		}
		return new Info(isBalanced, height);
	}

二、给定一棵二叉树的头节点head，任何两个节点之间都存在距离，返回整棵二叉树的最大距离

public static int maxDistance2(Node head) {
		return process(head).maxDistance;
	}

	public static class Info {
		public int maxDistance;
		public int height;

		public Info(int m, int h) {
			maxDistance = m;
			height = h;
		}

	}

	public static Info process(Node x) {
		if (x == null) {
			return new Info(0, 0);
		}
		Info leftInfo = process(x.left);
		Info rightInfo = process(x.right);
		int height = Math.max(leftInfo.height, rightInfo.height) + 1;
		int p1 = leftInfo.maxDistance;
		int p2 = rightInfo.maxDistance;
		int p3 = leftInfo.height + rightInfo.height + 1;
		int maxDistance = Math.max(Math.max(p1, p2), p3);
		return new Info(maxDistance, height);
	}

三、给定一棵二叉树的头节点head，返回这颗二叉树中最大的二叉搜索子树的头节点

public static class TreeNode {
		public int val;
		public TreeNode left;
		public TreeNode right;

		public TreeNode(int value) {
			val = value;
		}
	}

	// 提交如下的largestBSTSubtree方法，可以直接通过
	public static int largestBSTSubtree(TreeNode head) {
		if (head == null) {
			return 0;
		}
		return process(head).maxBSTSubtreeSize;
	}

	public static class Info {
		public int maxBSTSubtreeSize;
		public int allSize;
		public int max;
		public int min;

		public Info(int m, int a, int ma, int mi) {
			maxBSTSubtreeSize = m;
			allSize = a;
			max = ma;
			min = mi;
		}
	}

	public static Info process(TreeNode x) {
		if (x == null) {
			return null;
		}
		Info leftInfo = process(x.left);
		Info rightInfo = process(x.right);
		int max = x.val;
		int min = x.val;
		int allSize = 1;
		if (leftInfo != null) {
			max = Math.max(leftInfo.max, max);
			min = Math.min(leftInfo.min, min);
			allSize += leftInfo.allSize;
		}
		if (rightInfo != null) {
			max = Math.max(rightInfo.max, max);
			min = Math.min(rightInfo.min, min);
			allSize += rightInfo.allSize;
		}
		int p1 = -1;
		if (leftInfo != null) {
			p1 = leftInfo.maxBSTSubtreeSize;
		}
		int p2 = -1;
		if (rightInfo != null) {
			p2 = rightInfo.maxBSTSubtreeSize;
		}
		int p3 = -1;
		boolean leftBST = leftInfo == null ? true : (leftInfo.maxBSTSubtreeSize == leftInfo.allSize);
		boolean rightBST = rightInfo == null ? true : (rightInfo.maxBSTSubtreeSize == rightInfo.allSize);
		if (leftBST && rightBST) {
			boolean leftMaxLessX = leftInfo == null ? true : (leftInfo.max < x.val);
			boolean rightMinMoreX = rightInfo == null ? true : (x.val < rightInfo.min);
			if (leftMaxLessX && rightMinMoreX) {
				int leftSize = leftInfo == null ? 0 : leftInfo.allSize;
				int rightSize = rightInfo == null ? 0 : rightInfo.allSize;
				p3 = leftSize + rightSize + 1;
			}
		}
		return new Info(Math.max(p1, Math.max(p2, p3)), allSize, max, min);
	}

四、派对的最大快乐值

员工信息的定义如下:
class Employee {
public int happy; // 这名员工可以带来的快乐值
List subordinates; // 这名员工有哪些直接下级
}
派对的最大快乐值
公司的每个员工都符合 Employee 类的描述。整个公司的人员结构可以看作是一棵标准的、 没有环的多叉树。树的头节点是公司唯一的老板。除老板之外的每个员工都有唯一的直接上级。 叶节点是没有任何下属的基层员工(subordinates列表为空)，除基层员工外，每个员工都有一个或多个直接下级。

public static class Info {
		public int no;//直接下级不来的happy信息
		public int yes;//直接下级来的情况下happy的信息

		public Info(int n, int y) {
			no = n;
			yes = y;
		}
	}

	public static Info process(Employee x) {
		if (x == null) {
			return new Info(0, 0);
		}
		int no = 0;
		int yes = x.happy;
		for (Employee next : x.nexts) {
			Info nextInfo = process(next);
			no += Math.max(nextInfo.no, nextInfo.yes);
			yes += nextInfo.no;

		}
		return new Info(no, yes);
	}

五、二叉树的递归套路

1）假设以X节点为头，假设可以向X左树和X右树要任何信息
2）在上一步的假设下，讨论以X为头节点的树，得到答案的可能性（最重要）
3）列出所有可能性后，确定到底需要向左树和右树要什么样的信息
4）把左树信息和右树信息求全集，就是任何一棵子树都需要返回的信息S
5）递归函数都返回S，每一棵子树都这么要求
6）写代码，在代码中考虑如何把左树的信息和右树信息整合出整棵树的信息

可以通过x使用不使用来进行分情况处理。

六、给定一棵二叉树的头节点head，返回这颗二叉树是不是满二叉树

public static Info1 process1(Node head) {
		if (head == null) {
			return new Info1(0, 0);
		}
		Info1 leftInfo = process1(head.left);
		Info1 rightInfo = process1(head.right);
		int height = Math.max(leftInfo.height, rightInfo.height) + 1;
		int nodes = leftInfo.nodes + rightInfo.nodes + 1;
		return new Info1(height, nodes);
	}

七、给定一棵二叉树的头节点head，返回这颗二叉树中最大的二叉搜索子树的头节点

// 每一棵子树
	public static class Info {
		public Node maxSubBSTHead;
		public int maxSubBSTSize;
		public int min;
		public int max;

		public Info(Node h, int size, int mi, int ma) {
			maxSubBSTHead = h;
			maxSubBSTSize = size;
			min = mi;
			max = ma;
		}
	}

	public static Info process(Node X) {
		if (X == null) {
			return null;
		}
		Info leftInfo = process(X.left);
		Info rightInfo = process(X.right);
		int min = X.value;
		int max = X.value;
		Node maxSubBSTHead = null;
		int maxSubBSTSize = 0;
		if (leftInfo != null) {
			min = Math.min(min, leftInfo.min);
			max = Math.max(max, leftInfo.max);
			maxSubBSTHead = leftInfo.maxSubBSTHead;
			maxSubBSTSize = leftInfo.maxSubBSTSize;
		}
		if (rightInfo != null) {
			min = Math.min(min, rightInfo.min);
			max = Math.max(max, rightInfo.max);
			if (rightInfo.maxSubBSTSize > maxSubBSTSize) {
				maxSubBSTHead = rightInfo.maxSubBSTHead;
				maxSubBSTSize = rightInfo.maxSubBSTSize;
			}
		}
		if ((leftInfo == null ? true : (leftInfo.maxSubBSTHead == X.left && leftInfo.max < X.value))
				&& (rightInfo == null ? true : (rightInfo.maxSubBSTHead == X.right && rightInfo.min > X.value))) {
			maxSubBSTHead = X;
			maxSubBSTSize = (leftInfo == null ? 0 : leftInfo.maxSubBSTSize)
					+ (rightInfo == null ? 0 : rightInfo.maxSubBSTSize) + 1;
		}
		return new Info(maxSubBSTHead, maxSubBSTSize, min, max);
	}

八、给定一棵二叉树的头节点head，返回这颗二叉树中是不是完全二叉树

public static class TreeNode {
		public int val;
		public TreeNode left;
		public TreeNode right;

		public TreeNode(int v) {
			val = v;
		}
	}

	public static boolean isCompleteTree1(TreeNode head) {
		if (head == null) {
			return true;
		}
		LinkedList<TreeNode> queue = new LinkedList<>();
		// 是否遇到过左右两个孩子不双全的节点
		boolean leaf = false;
		TreeNode l = null;
		TreeNode r = null;
		queue.add(head);
		while (!queue.isEmpty()) {
			head = queue.poll();
			l = head.left;
			r = head.right;
			if (
			// 如果遇到了不双全的节点之后，又发现当前节点不是叶节点
			(leaf && (l != null || r != null)) || (l == null && r != null)

			) {
				return false;
			}
			if (l != null) {
				queue.add(l);
			}
			if (r != null) {
				queue.add(r);
			}
			if (l == null || r == null) {
				leaf = true;
			}
		}
		return true;
	}

方法二：

public static boolean isCompleteTree2(TreeNode head) {
		return process(head).isCBT;
	}

	public static class Info {
		public boolean isFull;
		public boolean isCBT;
		public int height;

		public Info(boolean full, boolean cbt, int h) {
			isFull = full;
			isCBT = cbt;
			height = h;
		}
	}

	public static Info process(TreeNode x) {
		if (x == null) {
			return new Info(true, true, 0);
		}
		Info leftInfo = process(x.left);
		Info rightInfo = process(x.right);
		int height = Math.max(leftInfo.height, rightInfo.height) + 1;
		boolean isFull = leftInfo.isFull && rightInfo.isFull && leftInfo.height == rightInfo.height;
		boolean isCBT = false;
		if (leftInfo.isFull && rightInfo.isFull && leftInfo.height == rightInfo.height) {
			isCBT = true;
		} else if (leftInfo.isCBT && rightInfo.isFull && leftInfo.height == rightInfo.height + 1) {
			isCBT = true;
		} else if (leftInfo.isFull && rightInfo.isFull && leftInfo.height == rightInfo.height + 1) {
			isCBT = true;
		} else if (leftInfo.isFull && rightInfo.isCBT && leftInfo.height == rightInfo.height) {
			isCBT = true;
		}
		return new Info(isFull, isCBT, height);
	}

九、给定一棵二叉树的头节点head，和另外两个节点a和b。返回a和b的最低公共祖先

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

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

	public static Node lowestAncestor1(Node head, Node o1, Node o2) {
		if (head == null) {
			return null;
		}
		// key的父节点是value
		HashMap<Node, Node> parentMap = new HashMap<>();
		parentMap.put(head, null);
		fillParentMap(head, parentMap);
		HashSet<Node> o1Set = new HashSet<>();
		Node cur = o1;
		o1Set.add(cur);
		while (parentMap.get(cur) != null) {
			cur = parentMap.get(cur);
			o1Set.add(cur);
		}
		cur = o2;
		while (!o1Set.contains(cur)) {
			cur = parentMap.get(cur);
		}
		return cur;
	}

	public static void fillParentMap(Node head, HashMap<Node, Node> parentMap) {
		if (head.left != null) {
			parentMap.put(head.left, head);
			fillParentMap(head.left, parentMap);
		}
		if (head.right != null) {
			parentMap.put(head.right, head);
			fillParentMap(head.right, parentMap);
		}
	}

方法二：

public static Node lowestAncestor2(Node head, Node a, Node b) {
		return process(head, a, b).ans;
	}

	public static class Info {
		public boolean findA;
		public boolean findB;
		public Node ans;

		public Info(boolean fA, boolean fB, Node an) {
			findA = fA;
			findB = fB;
			ans = an;
		}
	}

	public static Info process(Node x, Node a, Node b) {
		if (x == null) {
			return new Info(false, false, null);
		}
		Info leftInfo = process(x.left, a, b);
		Info rightInfo = process(x.right, a, b);
		boolean findA = (x == a) || leftInfo.findA || rightInfo.findA;
		boolean findB = (x == b) || leftInfo.findB || rightInfo.findB;
		Node ans = null;
		if (leftInfo.ans != null) {
			ans = leftInfo.ans;
		} else if (rightInfo.ans != null) {
			ans = rightInfo.ans;
		} else {
			if (findA && findB) {
				ans = x;
			}
		}
		return new Info(findA, findB, ans);
	}

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总结