初级5 二叉树

题目一：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

 递归版：递归来到每个节点的顺序

前序遍历  打印时机放在第1次来到这个节点的时候

 中序遍历  打印时机放在第2次来到这个节点的时候

 后序遍历  打印时机放在第3次来到这个节点的时候

    // 递归版本--先序遍历
    public static void preOrderRecur(Node head) {
        if (head == null) {
            return;
        }
        // System.out.print(head.value + " ");
        preOrderRecur(head.left);
        preOrderRecur(head.right);
    }

非递归版：stack栈

1,2，4，5，3，6,7

    // 非递归--前序遍历
    public static void preOrderUnRecur(Node head) {
        System.out.print("pre-order: ");
        if (head != null) {
            Stack<Node> stack = new Stack<Node>();
            stack.add(head);
            while (!stack.isEmpty()) {
                head = stack.pop();
                System.out.print(head.value + " ");
                
                //如果有右孩子，先压右
                // 如果有左孩子，后压左
                if (head.right != null) {
                    stack.push(head.right);
                }
                if (head.left != null) {
                    stack.push(head.left);
                }
            }
        }
        System.out.println();
    }

中左右被 模拟出来了

 先压右后压左

先弹左，后弹右

 

为什么不用队列？

递归函数就是一种栈

 

二叉树只有从上到下的路径，没有回去的路径

 非递归--中序遍历

    public static void inOrderUnRecur(Node head) {
        System.out.print("in-order: ");
        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;
                }
            }
        }
        System.out.println();
    }

head不等于空

 当前节点会把自己左边界全部压入栈中，head = head.left

head == null 停止

当前节点为空，从栈中拿出一个，打印，往右移动

当前节点不为空，当前压入栈中，节点往左移动

 极端例子

整棵树可以被左边界分解的

非递归--后序遍历

左右中  ---> 中右左----> 中左右(当前节点，先压左，再压右)   

先不打印，放入栈中，然后逆序输出

方法1：两个栈

    public static void posOrderUnRecur1(Node head) {
        System.out.print("pos-order: ");
        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);  打印时机，放入辅助栈中，
                if (head.left != null) {
                    s1.push(head.left);    // 左不为空，先压左
                } 
                if (head.right != null) {
                    s1.push(head.right);
                }
            }
            while (!s2.isEmpty()) {
                System.out.print(s2.pop().value + " ");   // 逆序打印栈
            }
        }
        System.out.println();
    }

 方法2:1个栈

纯属炫技

完整java代码

package class_04;

import java.util.Stack;

public class Code_01_PreInPosTraversal {

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

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


    // 递归版本--先序遍历
    public static void preOrderRecur(Node head) {
        if (head == null) {
            return;
        }
        // System.out.print(head.value + " ");
        preOrderRecur(head.left);
        preOrderRecur(head.right);
    }

    public static void inOrderRecur(Node head) {
        if (head == null) {
            return;
        }
        inOrderRecur(head.left);
        System.out.print(head.value + " ");
        inOrderRecur(head.right);
    }

    public static void posOrderRecur(Node head) {
        if (head == null) {
            return;
        }
        posOrderRecur(head.left);
        posOrderRecur(head.right);
        System.out.print(head.value + " ");
    }

    // 非递归--前序遍历
    public static void preOrderUnRecur(Node head) {
        System.out.print("pre-order: ");
        if (head != null) {
            Stack<Node> stack = new Stack<Node>();
            stack.add(head);
            while (!stack.isEmpty()) {
                head = stack.pop();
                System.out.print(head.value + " ");
                
                //如果有右孩子，先压右
                // 如果有左孩子，后压左
                if (head.right != null) {
                    stack.push(head.right);
                }
                if (head.left != null) {
                    stack.push(head.left);
                }
            }
        }
        System.out.println();
    }

    // 非递归---后序
    // 方法1：两个栈
    public static void inOrderUnRecur(Node head) {
        System.out.print("in-order: ");
        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;
                }
            }
        }
        System.out.println();
    }

    
    // 方法2：一个栈，炫技
    public static void posOrderUnRecur1(Node head) {
        System.out.print("pos-order: ");
        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);  打印时机，放入辅助栈中，
                if (head.left != null) {
                    s1.push(head.left);    // 左不为空，先压左
                } 
                if (head.right != null) {
                    s1.push(head.right);
                }
            }
            while (!s2.isEmpty()) {
                System.out.print(s2.pop().value + " ");   // 逆序打印栈
            }
        }
        System.out.println();
    }

    public static void posOrderUnRecur2(Node h) {
        System.out.print("pos-order: ");
        if (h != null) {
            Stack<Node> stack = new Stack<Node>();
            stack.push(h);
            Node c = null;
            while (!stack.isEmpty()) {
                c = stack.peek();
                if (c.left != null && h != c.left && h != c.right) {
                    stack.push(c.left);
                } else if (c.right != null && h != c.right) {
                    stack.push(c.right);
                } else {
                    System.out.print(stack.pop().value + " ");
                    h = c;
                }
            }
        }
        System.out.println();
    }

    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);

        // recursive
        System.out.println("==============recursive==============");
        System.out.print("pre-order: ");
        preOrderRecur(head);
        System.out.println();
        System.out.print("in-order: ");
        inOrderRecur(head);
        System.out.println();
        System.out.print("pos-order: ");
        posOrderRecur(head);
        System.out.println();

        // unrecursive
        System.out.println("============unrecursive=============");
        preOrderUnRecur(head);
        inOrderUnRecur(head);
        posOrderUnRecur1(head);
        posOrderUnRecur2(head);

    }

}

题目二：如何直观的打印一颗二叉树（福利）

 　　还原不了整棵树

package class_04;

public class Code_02_PrintBinaryTree {

    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);

    }

}

 

题目三：在二叉树中找到一个节点的后继节点

【题目】 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。

 新类型节点

 中序遍历中：

 问题描述：如果给了5，如何找到1，不用遍历整棵树

 

结论：

　　X 如果有右孩子，右子树

　　X 的后继节点一定是它右子树上的最左节点

原因是：中序 ： 左中右

X 没有右子树，

找他的parent，如果X是父的右子树，则继续

直到找到，X是父的左子树, 停止

则后继节点是parent

package class_04;

public class Code_03_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;
        }
        if (node.right != null) {
            return getLeftMost(node.right);  
        } else {    // 没有右子树
            Node parent = node.parent;
            while (parent != null && parent.left != node) {   // 有些节点没有后继，7的后继是Null
                node = parent;
                parent = node.parent;
            }
            return parent;   // 节点的后继
        }
    }

    public static Node getLeftMost(Node node) {  // 找到该子树的最左节点
        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));
    }

}

前驱节点

有左子树，左子树是最右节点

Node是当前父节点的右孩子，就停止

左神语录

面试5面， 2~3道题

迅速把面试官的题库挖空，还有吗，我要打10个

 

题目四：介绍二叉树的序列化和反序列化

data struct ----> 文件

方法1：_ #

序列化

_ 代表 分割， # 代表Null

 如何区分？

package class_04;

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

public class Code_04_SerializeAndReconstructTree {

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

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

    public static String serialByPre(Node head) {
        if (head == null) {
            return "#!";
        }
        String res = head.value + "!";
        res += serialByPre(head.left);   // 当前节点的左树返回个字符串
        res += serialByPre(head.right);   // 右树返回个字符串
        return res;
    }

    public static Node reconByPreString(String preStr) {
        String[] values = preStr.split("!");
        Queue<String> queue = new LinkedList<String>();
        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;
    }

    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;
    }

    public static Node reconByLevelString(String levelStr) {
        String[] values = levelStr.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;
    }

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

    // 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 = null;
        printTree(head);

        String pre = serialByPre(head);
        System.out.println("serialize tree by pre-order: " + pre);
        head = reconByPreString(pre);
        System.out.print("reconstruct tree by pre-order, ");
        printTree(head);

        String level = serialByLevel(head);
        System.out.println("serialize tree by level: " + level);
        head = reconByLevelString(level);
        System.out.print("reconstruct tree by level, ");
        printTree(head);

        System.out.println("====================================");

        head = new Node(1);
        printTree(head);

        pre = serialByPre(head);
        System.out.println("serialize tree by pre-order: " + pre);
        head = reconByPreString(pre);
        System.out.print("reconstruct tree by pre-order, ");
        printTree(head);

        level = serialByLevel(head);
        System.out.println("serialize tree by level: " + level);
        head = reconByLevelString(level);
        System.out.print("reconstruct tree by level, ");
        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("serialize tree by pre-order: " + pre);
        head = reconByPreString(pre);
        System.out.print("reconstruct tree by pre-order, ");
        printTree(head);

        level = serialByLevel(head);
        System.out.println("serialize tree by level: " + level);
        head = reconByLevelString(level);
        System.out.print("reconstruct tree by level, ");
        printTree(head);

        System.out.println("====================================");

        head = new Node(100);
        head.left = new Node(21);
        head.left.left = new Node(37);
        head.right = new Node(-42);
        head.right.left = new Node(0);
        head.right.right = new Node(666);
        printTree(head);

        pre = serialByPre(head);
        System.out.println("serialize tree by pre-order: " + pre);
        head = reconByPreString(pre);
        System.out.print("reconstruct tree by pre-order, ");
        printTree(head);

        level = serialByLevel(head);
        System.out.println("serialize tree by level: " + level);
        head = reconByLevelString(level);
        System.out.print("reconstruct tree by level, ");
        printTree(head);

        System.out.println("====================================");

    }
}

 反序列化

 

 方法2：按层序列化

题目六：判断一棵二叉树是否是平衡二叉树