【算法编程】二叉树经典题（基础篇）

二叉树的遍历

  二叉树遍历分为三种：前序、中序和后序，为什么这么命名呢？其实是根据节点顺序命名的。

如图为满节点，1为根节点、2为做节点、3为右节点，主要是看根结点1的位置，在前面就是前序遍历、在中间就是中序遍历、在后面就是后续遍历。

下面我们通过例子，来看看：

1、前序遍历

• 在遍历节点的左节点的时候，可能左边节点还有左节点和右节点，要一直遍历到叶子节点，才能遍历其右节点
• 例如在遍历到节点2时，它的左节点是3，但是节点3不是叶子节点，此时就不能去遍历节点2的右节点4
• 虽然节点3没有左节点，但是有右节点5，它应该在节点2的右节点4的前面
• 也就是我们在遍历的时候，要将左边节点遍历完才能遍历右边节点

2、中序遍历

3、后序遍历

一、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

1、递归版

package day03;

/**
 * 二叉树的遍历（递归版）
 * @author Danycym
 *
 */
public class Code01_PreInPosTraversalRecur {
	
	//树节点
	public static class Node{
		public int value;
		public Node left;
		public Node right;
		
		public Node(int data) {
			this.value = data;
		}
	}
	
	//前序遍历
	public static void preOrderRecur(Node head) {
		if(head == null) {
			return;
		}
		System.out.print(head.value + " ");
		preOrderRecur(head.left);
		preOrderRecur(head.right);
	}
	
	//中序遍历
	public static void inOrderRecur(Node head) {
		if(head == null) {
			return;
		}
		inOrderRecur(head.left);
		System.out.print(head.value + " ");
		inOrderRecur(head.right);
	}
	
	//后序遍历
	public static void posOrderRecur(Node head) {
		if(head == null) {
			return;
		}
		posOrderRecur(head.left);			
		posOrderRecur(head.right);
		System.out.print(head.value + " ");
	}
	
	public static void main(String[] args) {
		Node head = new Node(5);
		head.left = new Node(3);
		head.right = new Node(8);
		head.left.left = new Node(2);
		head.left.right = new Node(4);
		head.left.left.left = new Node(1);
		head.right.left = new Node(7);
		head.right.left.left = new Node(6);
		head.right.right = new Node(10);
		head.right.right.left = new Node(9);
		head.right.right.right = new Node(11);
		
		System.out.print("前序遍历：");
		preOrderRecur(head);
		System.out.print("\n中序遍历：");
		inOrderRecur(head);
		System.out.print("\n后序遍历：");
		posOrderRecur(head);
	}
}

2、非递归版图解

上图没执行到最后，后面的按照重复模式执行下去即可

3、Java代码：

package day03;

import java.util.Stack;

/**
 * 二叉树遍历（非递归版）
 * @author Danycym
 *
 */
public class Code02_PreInPosTraversalUnRecur {
	public static class Node{
		public int value;
		public Node left;
		public Node right;
		
		public Node(int data) {
			this.value = data;
		}
	}

	//前序遍历
	public static void preOrderUnRecur(Node head) {
		if(head != null) {
			Stack<Node> stack = new Stack<Node>();
			stack.push(head);
			while(!stack.isEmpty()) {
				head = stack.pop();
				System.out.print(head.value + " ");
				if(head.right != null) {
					stack.push(head.right);
				}
				if(head.left != null) {
					stack.push(head.left);
				}
			}
		}
	}
	
	//中序遍历
	public static void inOrderUnRecur(Node head) {
		if(head != null) {
			Stack<Node> stack = new Stack<Node>();
			while(!stack.isEmpty() || head != null) {
				if(head != null) {
					stack.push(head);
					head = head.left;
				}else {
					head = stack.pop();
					System.out.print(head.value + " ");
					head = head.right;
				}
			}
		}
	}
	
	//后序遍历1
	public static void posOrderUnRecur1(Node head) {
		if(head != null) {
			Stack<Node> stack = new Stack<Node>();
			stack.push(head);
			Node c = null;
			while(!stack.isEmpty()) {
				c = stack.peek();
				if(c.left != null && head != c.left && head != c.right) {
					stack.push(c.left);
				}else if(c.right != null && head != c.right) {
					stack.push(c.right);
				}else {
					System.out.print(stack.pop().value + " ");
					head = c;
				}
			}
		}
	}
	
	//后序遍历2
	public static void posOrderUnRecur2(Node head) {
		if(head != null) {
			Stack<Node> s1 = new Stack<Node>();
			Stack<Node> s2 = new Stack<Node>();
			s1.push(head);
			while(!s1.isEmpty()) {
				head = s1.pop();
				s2.push(head);
				if(head.left != null) {
					s1.push(head.left);
				}
				if(head.right != null) {
					s1.push(head.right);
				}
			}
			
			while(!s2.isEmpty()) {
				System.out.print(s2.pop().value + " ");
			}
		}
	}
	
	
	
	public static void main(String[] args) {
		Node head = new Node(5);
		head.left = new Node(3);
		head.right = new Node(8);
		head.left.left = new Node(2);
		head.left.right = new Node(4);
		head.left.left.left = new Node(1);
		head.right.left = new Node(7);
		head.right.left.left = new Node(6);
		head.right.right = new Node(10);
		head.right.right.left = new Node(9);
		head.right.right.right = new Node(11);
		
		System.out.print("前序遍历：");
		preOrderUnRecur(head);
		System.out.print("\n中序遍历：");
		inOrderUnRecur(head);
		System.out.print("\n后序遍历1：");
		posOrderUnRecur1(head);
		System.out.print("\n后序遍历2：");
		posOrderUnRecur2(head);

	}

}

二、在二叉树中找到一个节点的后继节点

1、题目描述

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。

2、图解

3、Java代码

package day03;

public class Code03_SuccessorNode {
	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;
		}
		//有右子树
		if(node.right != null) {
			return getLeftMost(node.right);  //获取右子树最左的节点
		}else {//没有右子树
			Node parent = node.parent;  //取父节点
			while(parent != null && parent.left != node) {  //父节点不为空，求当前节点不是父节点的左节点
				node = parent;  //父节点设置为当前节点
				parent = node.parent;  //再找父节点的父节点
			}
			return parent;
		}
	}
	
	//获取右子树的最左的节点
	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(1);
		head.parent = null;
		head.left = new Node(2);
		head.left.parent = head;
		head.right = new Node(3);
		head.right.parent = head;
		head.left.left = new Node(4);
		head.left.left.parent = head.left;
		head.left.right = new Node(5);
		head.left.right.parent = head.left;
		head.right.left = new Node(6);
		head.right.left.parent = head.right;
		head.right.right = new Node(7);
		head.right.right.parent = head.right;
		head.left.right.left = new Node(8);
		head.left.right.left.parent = head.left.right;
		head.right.right.left = new Node(9);
		head.right.right.left.parent = head.right.right;
		head.right.right.right = new Node(10);
		head.right.right.right.parent = head.right.right;
		
		Node test = head;
		System.out.println(test.value + "  next:  " + getSuccessorNode(test).value);
		test = head.left;
		System.out.println(test.value + "  next:  " + getSuccessorNode(test).value);
		test = head.right;
		System.out.println(test.value + "  next:  " + getSuccessorNode(test).value);
		test = head.left.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.right.left;
		System.out.println(test.value + "  next:  " + getSuccessorNode(test).value);
		test = head.right.right;
		System.out.println(test.value + "  next:  " + getSuccessorNode(test).value);
		test = head.left.right.left;
		System.out.println(test.value + "  next:  " + getSuccessorNode(test).value);
		test = head.right.right.left;
		System.out.println(test.value + "  next:  " + getSuccessorNode(test).value);
		test = head.right.right.right;
		System.out.println(test.value + "  next:  " + getSuccessorNode(test));
	}

}

三、二叉树的序列化和反序列化

1、图解

2、Java代码

package day03;

import java.util.LinkedList;
import java.util.Queue;

public class Code04_SerializeAndReconstructTree {
	public static class Node{
		public int value;
		public Node left;
		public Node right;
		
		public Node(int data) {
			this.value = data;
		}
	}
	
	//二叉树按前序遍历序列化
	public static String serialByPre(Node head) {
		if(head == null) {
			return "#!";
		}
		String res = head.value + "!";
		res += serialByPre(head.left);
		res += serialByPre(head.right);
		return res;
	}
	//前序遍历反序列化
	public static Node reconByPreString(String str) {
		String[] values = str.split("!");  //按"!"拆分
		Queue<String> queue = new LinkedList<String>();  //队列
		for(int i = 0; i != values.length; i++) {
			queue.offer(values[i]);
		}
		return reconPreOrder(queue);
	}
	public static Node reconPreOrder(Queue<String> queue) {
		String value = queue.poll();
		if(value.equals("#")) {
			return null;
		}
		Node head = new Node(Integer.valueOf(value));
		head.left = reconPreOrder(queue);
		head.right = reconPreOrder(queue);
		return head;
	}
	
	
	
	
	//层次遍历序列化
	public static String serialByLevel(Node head) {
		if(head == null) {
			return "#!";
		}
		String res = head.value + "!";
		Queue<Node> queue = new LinkedList<Node>();
		queue.offer(head);
		while(!queue.isEmpty()) {
			head = queue.poll();
			if(head.left != null) {
				res += head.left.value + "!";
				queue.offer(head.left);
			}else {
				res += "#!";
			}
			
			if(head.right != null) {
				res += head.right.value + "!";
				queue.offer(head.right);
			}else {
				res += "#!";
			}
		}
		return res;
	}
	
	//层次遍历反序列化
	public static Node reconByLevelString(String str) {
		String[] values = str.split("!");
		int index = 0;
		Node head = generateNodeByString(values[index++]);
		Queue<Node> queue = new LinkedList<Node>();
		if(head != null) {
			queue.offer(head);
		}
		Node node = null;
		while(!queue.isEmpty()) {
			node = queue.poll();
			node.left = generateNodeByString(values[index++]);
			node.right = generateNodeByString(values[index++]);
			if(node.left != null) {
				queue.offer(node.left);
			}
			if(node.right != null) {
				queue.offer(node.right);
			}
		}
		return head;
	}
	

	private static Node generateNodeByString(String val) {
		if(val.equals("#")) {
			return null;
		}
		return new Node(Integer.valueOf(val));
	}

	//打印二叉树结构
	public static void printTree(Node head) {
		System.out.println("二叉树：");
		printInOrder(head, 0, "H", 17);
	}

	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 = null;
		printTree(head);
		
		String pre = serialByPre(head);
		System.out.println("前序遍历序列化: " + pre);
		head = reconByPreString(pre);
		System.out.println("前序遍历反序列化:");
		printTree(head);
		System.out.println("====================================");
		
		head = new Node(1);
		printTree(head);
		
		pre = serialByPre(head);
		System.out.println("前序遍历序列化: " + pre);
		head = reconByPreString(pre);
		System.out.println("前序遍历反序列化:");
		printTree(head);
		System.out.println("====================================");
		
		head = new Node(1);
		head.left = new Node(2);
		head.right = new Node(3);
		head.left.left = new Node(4);
		head.right.right = new Node(5);
		printTree(head);
		
		pre = serialByPre(head);
		System.out.println("前序遍历序列化: " + pre);
		head = reconByPreString(pre);
		System.out.println("前序遍历反序列化:");
		printTree(head);
		System.out.println("====================================");
		
		head = new Node(1);
		head.left = new Node(2);
		head.left.left = new Node(4);
		head.left.right = new Node(5);
		head.right = new Node(3);
		head.right.left = new Node(6);
		head.right.right = new Node(7);
		printTree(head);
		
		pre = serialByPre(head);
		System.out.println("前序遍历序列化: " + pre);
		head = reconByPreString(pre);
		System.out.println("前序遍历反序列化:");
		printTree(head);
		System.out.println("====================================");
		
		
		head = new Node(1);
		head.left = new Node(2);
		head.left.left = new Node(4);
		head.left.right = new Node(5);
		head.right = new Node(3);
		head.right.left = new Node(6);
		head.right.right = new Node(7);
		printTree(head);
		
		pre = serialByLevel(head);
		System.out.println("层次遍历序列化: " + pre);
		head = reconByLevelString(pre);
		System.out.println("层次遍历反序列化:");
		printTree(head);
		System.out.println("====================================");	
	}

}

四、折纸问题

1、题目描述

  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

2、图解

程序中，可以通过一个 boolean 类型的变量表示是折痕的左边还是右边

3、Java代码

package day03;

public class Code05_PaperFolding {

	public static void printAllFolds(int N) {
		printProcess(1, N, true);
	}

	public static void printProcess(int i, int N, boolean down) {
		if (i > N) {
			return;
		}
		printProcess(i + 1, N, true);
		System.out.print(down ? "down " : "up ");
		printProcess(i + 1, N, false);
	}

	public static void main(String[] args) {
		int N = 1;
		printAllFolds(N);
		System.out.println();
		N = 2;
		printAllFolds(N);
		System.out.println();
		N = 3;
		printAllFolds(N);
	}
}

五、判断一棵二叉树是否是平衡二叉树

• 平衡二叉树：对于二叉树中的任意一个节点，它的左子树和右子树的高度差不大于1
• 满二叉树一定是平衡二叉树

1、解题思路

判断一棵二叉树是否是平衡二叉树，就需要判断每个节点的左子树和右子树是否平衡，且左子树与右子树高度差

• 1）左子树是否平衡
• 2）右子树是否平衡
• 3）左子树高度
• 4）右子树高度

那么可以通过递归的方式实现

2、Java代码

package day03;

/**
 * 判断是否为平衡二叉树
 * @author Danycym
 *
 */
public class Code06_IsBalanceTree {
	public static class Node{
		public int value;
		public Node left;
		public Node right;
		
		public Node(int data) {
			this.value = data;
		}
	}
	
	//第一种写法
	//返回结构
	public static class ReturnData{
		public boolean isB;
		public int h;

		public ReturnData(boolean isB, int h){
			this.isB = isB;
			this.h = h;
		}
	}
	//主函数
	public static boolean isB(Node head) {
		return process(head).isB;
	}
	//递归函数
	public static ReturnData process(Node head){
		if(head == null){
			return new ReturnData(true, 0);
		}
		ReturnData leftData = process(head.left);
		if(!leftData.isB){
			return new ReturnData(false, 0);
		}
		ReturnData rightData = process(head.right);
		if(!rightData.isB){
			return new ReturnData(false, 0);
		}
		if(Math.abs(leftData.h - rightData.h) > 1){
			return new ReturnData(false, 0);
		}
		return new ReturnData(true, Math.max(leftData.h , rightData.h) + 1);
	}
	
	//第二种写法
	//主函数
	public static boolean isBalance(Node head) {
		boolean[] res = new boolean[1];
		res[0] = true;
		getHeight(head, 1, res);
		return res[0];
	}
	//求树高度
	public static int getHeight(Node head, int level, boolean[] res) {
		if (head == null) {
			return level;
		}
		int lH = getHeight(head.left, level + 1, res);
		if (!res[0]) {
			return level;
		}
		int rH = getHeight(head.right, level + 1, res);
		if (!res[0]) {
			return level;
		}
		if (Math.abs(lH - rH) > 1) {
			res[0] = false;
		}
		return Math.max(lH, rH);
	}
	
	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);
		
		System.out.println(isB(head));
		System.out.println(isBalance(head));
		
		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.left.left.left = new Node(6);
		
		System.out.println(isB(head));
		System.out.println(isBalance(head));

	}
}

六、判断一棵树是否是搜索二叉树

• 搜索二叉树：二叉树中的任意一个节点，它的左子树都比它小，右子树都比它大

1、解题思路

• 我们只需要进行中序遍历，看中序遍历是否是从小到大排序的即可

2、Java代码

package day03;

import java.util.Stack;

public class Code07_IsBST {
	public static class Node{
		public int value;
		public Node left;
		public Node right;
		
		public Node(int data) {
			this.value = data;
		}
	}

	public static boolean isBST(Node head) {
		int pre = Integer.MIN_VALUE;
		if(head != null) {
			Stack<Node> stack = new Stack<Node>();
			while(!stack.isEmpty() || head != null) {
				if(head != null) {
					stack.push(head);
					head = head.left;
				}else {
					head = stack.pop();
					if(head.value < pre) {
						return false;
					}
					pre = head.value;
					head = head.right;
				}
			}
		}
		return true;
	}
	
	//打印二叉树 
	public static void printTree(Node head) {
		System.out.println("二叉树为:");
		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(4);
		head.left = new Node(2);
		head.right = new Node(6);
		head.left.left = new Node(1);
		head.left.right = new Node(3);
		head.right.left = new Node(5);

		printTree(head);
		System.out.println(isBST(head));

	}
}

七、判断一棵树是否是完全二叉树

• 完全二叉树：如果一棵二叉树深度为k，则除第k层外其余所有层的所有节点都有左节点和右节点（即度为2），且叶子节点从左到右依次存在。

1、解题思路

情况可总结如下：

• 1）如果一个节点有右孩子，没有左孩子，肯定不是完全二叉树
• 2）如果一个节点有左没右或者左右都没有，那么它后面遍历到的节点必须是叶节点，否则肯定不是完全二叉树

按层遍历

2、Java代码

package day03;

import java.util.LinkedList;
import java.util.Queue;

public class Code08_IsCBT {
	public static class Node {
		public int value;
		public Node left;
		public Node right;

		public Node(int data) {
			this.value = data;
		}
	}
	public static boolean isCBT(Node head) {
		if (head == null) {
			return true;
		}
		Queue<Node> queue = new LinkedList<Node>();
		boolean leaf = false;
		Node l = null;
		Node r = null;
		queue.offer(head);
		while (!queue.isEmpty()) {
			head = queue.poll();
			l = head.left;
			r = head.right;
			//(leaf && (l != null || r != null)) 叶节点，左孩子或右孩子不为空，返回false
			//(l == null && r != null) 情况一，左孩子为空，右孩子不为空
			if ((leaf && (l != null || r != null)) || (l == null && r != null)) {
				return false;
			}
			if (l != null) {
				queue.offer(l);
			}
			if (r != null) {
				queue.offer(r);
			}else{  //右孩子为空，开启叶子节点，也就是后面遍历到的节点要是叶子节点
				leaf = true;
			}
		}
		return true;
	}

	// for test -- print tree
	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(4);
		head.left = new Node(2);
		head.right = new Node(6);
		head.left.left = new Node(1);
		head.left.right = new Node(3);
		head.right.left = new Node(5);

		printTree(head);
		System.out.println(isCBT(head));
		
		head = new Node(4);
		head.left = new Node(2);
		head.right = new Node(6);
		head.left.left = new Node(1);
		head.left.right = new Node(3);
		head.right.right = new Node(5);

		printTree(head);
		System.out.println(isCBT(head));

	}
}

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
• 按要求我们就不能直接遍历二叉树

1、解题思路

• 完全二叉树是由满二叉树而引出来的
• 一棵满二叉树，高度为 $L$ ，那么整棵树的节点个数为 $2^L-1$
• 那么我们就可以：
• 1）先遍历完全二叉树的左边界，可以记录完全二叉树的高度h
• 2）再遍历完全二叉树头结点的右子树的左边界，看它到没到完全二叉树的最高高度h，如果达到最高高度h，那么头结点的左子树就是满节点，为 $2^{(h-1)}-1$ 个，再加上头结点为 $2^{(h-1)}-1+1$，头结点的右子树再进行递归
• 3）如果头结点的右子树没有达到完全二叉树的最高高度h， 那么头结点的右子树就是满节点，只是高度右子树的高度比头结点的左子树的高度少1，那么节点个数为 $2^{(h-2)}-1$，再加上头结点为 $2^{(h-2)}-1+1$，头结点的左子树再进行递归

2、Java代码

package day03;

public class Code09_CompleteTreeNodeNumber {

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

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

	public static int nodeNum(Node head) {
		if (head == null) {
			return 0;
		}
		return bs(head, 1, mostLeftLevel(head, 1));
	}
	/**
	 * 
	 * @param node  当前节点
	 * @param level  当前节点在第几层
	 * @param h  树的高度，固定值
	 * @return  返回节点个数
	 */
	public static int bs(Node node, int level, int h) {
		if (level == h) {
			return 1;
		}
		if (mostLeftLevel(node.right, level + 1) == h) {
			return (1 << (h - level)) + bs(node.right, level + 1, h);
		} else {
			return (1 << (h - level - 1)) + bs(node.left, level + 1, h);
		}
	}
	
	/**
	 * 求二叉树的高度
	 * @param node  当前节点
	 * @param level  所在层数
	 * @return  返回二叉树高度
	 */
	public static int mostLeftLevel(Node node, int level) {
		while (node != null) {
			level++;
			node = node.left;
		}
		return level - 1;
	}

	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);
		System.out.println(nodeNum(head));

	}

}