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切换日光/暗黑模式

二叉树的套路 ​

1. 判断二叉树是否是完全二叉树 ​

1.1 方法一：层序遍历 ​

  • 层序遍历
  • 若某节点有右孩子，无左孩子，一定不是完全二叉树
  • 若第一次遇到某节点，其左右节点为NULL，剩下的节点一定是叶子节点
java
	public static class Node {
		public int value;
		public Node left;
		public Node right;

		public Node(int data) {
			this.value = data;
		}
	}
java
	public static boolean isCBT1(Node head) {
		if (head == null) {
			return true;
		}
		LinkedList<Node> queue = new LinkedList<>();
		// 是否遇到过左右两个孩子不双全的节点
		boolean leaf = false;
		Node l = null;
		Node 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);
			}
            // 当遇到非全孩子时，之后的每一个节点一定是叶节点
			if (l == null || r == null) {
				leaf = true;
			}
		}
		return true;
	}

1.2 方法二：模板 ​

  • 左满 右满 左高 == 右高
  • 左完 右满 左高 == 右高 +1
  • 左满 右满 左高 == 右高 +1
  • 左满 右完 左高 == 右高
java
	public static class Node {
		public int value;
		public Node left;
		public Node right;

		public Node(int data) {
			this.value = data;
		}
	}
    public static class Info {
		public boolean isFull;   // 是否为满二叉树
		public boolean isCBT;    // 是否是完全二叉树
		public int height;

		public Info(boolean full, boolean cbt, int h) {
			isFull = full;
			isCBT = cbt;
			height = h;
		}
	}
java
	public static Info process(Node 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;
        // 当前X是否为满二叉树， 左满 右满 且 高度一样
		boolean isFull = leftInfo.isFull && rightInfo.isFull && leftInfo.height == rightInfo.height;
        // 是否完全二叉树
		boolean isCBT = false;
        
		if (leftInfo.isFull && rightInfo.isFull && leftInfo.height == rightInfo.height) {
			isCBT = true;  // 左满   右满   左高  == 右高
		} else if (leftInfo.isCBT && rightInfo.isFull && leftInfo.height == rightInfo.height + 1) { 
			isCBT = true;  // 左完   右满   左高 == 右高 +1
		} else if (leftInfo.isFull && rightInfo.isFull && leftInfo.height == rightInfo.height + 1) {
			isCBT = true;  // 左满   右满   左高 == 右高 +1 
		} else if (leftInfo.isFull && rightInfo.isCBT && leftInfo.height == rightInfo.height) {
			isCBT = true;  // 左满   右完   左高  == 右高
		}
		return new Info(isFull, isCBT, height);
	}
java
	public static boolean isCBT2(Node head) {
		return process(head).isCBT;
    }

2. 给定一颗二叉树的头结点，判断这棵树是否是平衡二叉树 ​

  • X的左树平衡
  • X的右树平衡
  • |X左 - X右| < 2
java
	public static class Node {
		public int value;
		public Node left;
		public Node right;

		public Node(int data) {
			this.value = data;
		}
	}
java
	public static class Info{
		public boolean isBalanced;
		public int height;
		
		public Info(boolean i, int h) {
			isBalanced = i;
			height = h;
		}
	}
java
	public static Info process(Node x) {
		if(x == null) {
			return new Info(true, 0);
		}// 空树一定是
		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) {  // 左树不平
			isBalanced = false;
		}
		if(!rightInfo.isBalanced) { 
			isBalanced = false;
		}
		if(Math.abs(leftInfo.height - rightInfo.height) > 1) {
			isBalanced = false;
		}
		return new Info(isBalanced, height);
	}
java
	// 主函数
    public static boolean isBalanced2(Node head) {
		return process(head).isBalanced;
	}

3. 判断是否是搜索二叉树 ​

  • X 左边是搜索二叉树
  • X 右边是搜索二叉树
  • X 左边的最大值
  • X 右边的最小值
java
	public static class Node {
		public int value;
		public Node left;
		public Node right;

		public Node(int data) {
			this.value = data;
		}
	}
java
	public static class Info {
		public boolean isBST;
		public int max;
		public int min;

		public Info(boolean i, int ma, int mi) {
			isBST = i;
			max = ma;
			min = mi;
		}
	}
java
	public static Info process(Node x) {
        // 空树 上游处理
		if (x == null) { 
			return null;
		}
        // 向左数 和 右树要信息
		Info leftInfo = process(x.left);
		Info rightInfo = process(x.right);
        
		int max = x.value;
		if (leftInfo != null) {  //有左树，那么就PK下
			max = Math.max(max, leftInfo.max);
		}
		if (rightInfo != null) {
			max = Math.max(max, rightInfo.max);
		}
        
		int min = x.value;
		if (leftInfo != null) {
			min = Math.min(min, leftInfo.min);
		}
		if (rightInfo != null) {
			min = Math.min(min, rightInfo.min);
		}
        
		boolean isBST = true;
		if (leftInfo != null && !leftInfo.isBST) {
			isBST = false;  //左树不为空  且  左树不是线索二叉树，违规
		}
		if (rightInfo != null && !rightInfo.isBST) {
			isBST = false;
		}
		if (leftInfo != null && leftInfo.max >= x.value) {
			isBST = false;  // 左树不为空 且 左树的最大值大于当前值，违规
		}
		if (rightInfo != null && rightInfo.min <= x.value) {
			isBST = false;
		}
		return new Info(isBST, max, min);
	}
java
    // 主函数
	public static boolean isBST2(Node head) {
		if (head == null) { // 上游处理
			return true;
		}
		return process(head).isBST;
	}

4. 任何两个节点存在距离，求整颗二叉树的最大距离 ​

  • X 左树上有最大距离，与X无关
  • X 右树上有最大距离，与X无关
  • 经过X，左树离X最远+右树离X最远+1 （左树高度+右树高度+1）

要求：最大dis 树高

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

		public Node(int data) {
			this.value = data;
		}
	}
java
	public static class Info {
		public int maxDistance;
		public int height;

		public Info(int m, int h) {
			maxDistance = m;
			height = h;
		}
	}
java
	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);
	}
java
    public static int maxDistance2(Node head) {
        return process(head).maxDistance;
    }

5. 判断一棵树是否是满二叉树 ​

  • 左树是否满二叉树
  • 右树是否满二叉树
  • 左树和右树高度是否一样
java
	public static class Node {
		public int value;
		public Node left;
		public Node right;

		public Node(int data) {
			this.value = data;
		}
	}
java
	public static class Info2 {
		public boolean isFull;
		public int height;

		public Info2(boolean f, int h) {
			isFull = f;
			height = h;
		}
	}
java
	public static Info2 process2(Node h) {
		if (h == null) {
			return new Info2(true, 0);
		}
		Info2 leftInfo = process2(h.left);
		Info2 rightInfo = process2(h.right);
		boolean isFull = leftInfo.isFull && rightInfo.isFull && leftInfo.height == rightInfo.height;
		int height = Math.max(leftInfo.height, rightInfo.height) + 1;
		return new Info2(isFull, height);
	}
java
	public static boolean isFull2(Node head) {
		if (head == null) {
			return true;
		}
		return process2(head).isFull;
	}

6. 整颗二叉树不是线索二叉树，求子树是搜索二叉树，且节点最多有多少 ​

子树：包括当前节点，底下全要

  • X不做头，左边有MaxSize
  • X不做头，右边有MaxSize
  • X做头，左树和右树是搜索二叉树，且左树最大值<X，右树最小值>X，最后左边+右边+1
java
	public static class Node {
		public int value;
		public Node left;
		public Node right;

		public Node(int data) {
			this.value = data;
		}
	}
java
	public static class Info {
		public int maxBSTSubtreeSize; //最大子树的Size
		public int allSize;   //整棵树的size
		public int max;
		public int min;

		public Info(int m, int a, int ma, int mi) {
			maxBSTSubtreeSize = m;
			allSize = a;
			max = ma;
			min = mi;
		}
	}
java
	public static Info process(Node x) {
		if (x == null) {
			return null;
		}
		Info leftInfo = process(x.left);
		Info rightInfo = process(x.right);
        
		int max = x.value;
		int min = x.value;
		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; //如果左树为空，那么-1
		if (leftInfo != null) {
			p1 = leftInfo.maxBSTSubtreeSize;
		}
		int p2 = -1;
		if (rightInfo != null) {
			p2 = rightInfo.maxBSTSubtreeSize;
		}
        
		int p3 = -1;
        // 判断左树是不是搜索二叉树
		boolean leftBST = leftInfo == null ? true : (leftInfo.maxBSTSubtreeSize == leftInfo.allSize);
		boolean rightBST = rightInfo == null ? true : (rightInfo.maxBSTSubtreeSize == rightInfo.allSize);
        
		if (leftBST && rightBST) {
            // 判断当前节点是否大于左树的最大值
			boolean leftMaxLessX = leftInfo == null ? true : (leftInfo.max < x.value);
			boolean rightMinMoreX = rightInfo == null ? true : (x.value < 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);
	}
java
	public static int maxSubBSTSize2(Node head) {
		if(head == null) {
			return 0;
		}
		return process(head).maxBSTSubtreeSize;
	}

7. 返回二叉树中最大的二叉搜索子树的头节点 ​

java
	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;
		}
	}
java
	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);
	}
java
	public static Node maxSubBSTHead2(Node head) {
		if (head == null) {
			return null;
		}
		return process(head).maxSubBSTHead;
	}

8. 返回A和B的最近公共祖先 ​

8.1 方法1：模板 ​

  • X无关，左树上有答案，右树上有答案，不全
  • X有关，
java
	public static class Info {
		public boolean findA;
		public boolean findB;
		public Node ans;      //最终结果

		public Info(boolean fA, boolean fB, Node an) {
			findA = fA;
			findB = fB;
			ans = an;
		}
	}
java
	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;  // 左树已经找到了
		} else if (rightInfo.ans != null) {
			ans = rightInfo.ans; // 右树已经找到了
		} else {
			if (findA && findB) {
				ans = x;      //  左树和右树同时找到，当前节点为结果
			}
		}
		return new Info(findA, findB, ans);
	}

8.2 方法二：Leetcode上 ​

9. 公司问题 ​

一颗多叉树，可以选若干节点参加聚会

直接上下级不能同时到场

问：如何让Happy值最大

java
	public static class Employee {
		public int happy;
		public List<Employee> nexts;

		public Employee(int h) {
			happy = h;
			nexts = new ArrayList<>();
		}
	}
  • X 来的情况下，X.happy + X.children不来的max
  • X 不来的情况下，max
java
	public static class Info {
		public int no;          // 整颗子树不来的最大收益
		public int yes;

		public Info(int n, int y) {
			no = n;
			yes = y;
		}
	}
java
	public static Info process(Employee x) {
		if (x == null) {
			return new Info(0, 0);
		}
		int no = 0;
		int yes = x.happy;  // 来的话就是当前x的快乐值
        
		for (Employee next : x.nexts) {
			Info nextInfo = process(next);
            // 如果X不来，子代可来可不来
			no += Math.max(nextInfo.no, nextInfo.yes);
            // 如果X来，子代一定不来
			yes += nextInfo.no;

		}
		return new Info(no, yes);
	}
java
	public static int maxHappy2(Employee head) {
		Info allInfo = process(head);
        // 要么最大老板不来 或 来 的收益
		return Math.max(allInfo.no, allInfo.yes);
	}