牛客网初级算法之四

题目1:前序，中序，后序 非递归遍历二叉树

代码：

前序：

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 + " ");
                if (head.right != null) {
                    stack.push(head.right);
                }
                if (head.left != null) {
                    stack.push(head.left);
                }
            }
        }
        System.out.println();
    }

中序：

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();
    }

后序

最简单的思路：

前序的顺序为 中左右 后序的顺序为 左右中，若是前序遍历变为 中右左，那么后序遍历和前序遍历的顺序就会正好相反，如何让前序遍历 变为 中右左呢，就是把left和right入站的顺序

改变一下就行，而后在每次打印节点的地方，换成把节点加入到另外一个栈，以后遍历这个站就是 后序遍历的节点顺序

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);
                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 + " ");
            }
        }
        System.out.println();
    }

题目二：

二叉树的序列号和反序列化

思路一：利用递归方法：

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;
    }

反序列化：

怎么序列化的就怎么反序列化：

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;
    }

思路二：利用树的层次

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));
    }

题目三：求二叉树的后继节点（每一个节点有parent指针，指向父亲节点）

后继节点是指二叉树中序遍历中，某个节点的后面的那个节点

思路：

经过概括总结，可分为两种状况

状况1:节点A的右子树不为空，那么节点A的后继为 右子树的 最左边的孩子

状况2:节点A的右子树为空，那么开始从节点A不断经过parent指针找到他的父节点，直到找到的 当前节点A' 的父亲节点B的左孩子等于A',B就是节点A的后继

特殊状况为 中序遍历的最后一个节点，它没有后继，返回null

代码以下：

public class 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 {
            Node parent = node.parent;
            while (parent != null && parent.left != node) {
                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));
    }

}

题目四：判断一个树是否是平衡二叉树？

平衡二叉树是指任何树的左子树和右子树的高度相差不大于1

思路

根据概念能够分析出能够用递归方法：

1 求出节点A的 左子树高度

2 求出节点A的 右子树高度

3 判断高度差是否大于1，更新布尔值

public class IsBalancedTree {

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

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

    public static boolean isBalance(Node head) {
        boolean[] res = new boolean[1];
        res[0] = true;
        getHeight(head, 1, res);
        return res[0];
    }

    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(isBalance(head));

    }

}

这里经过层级字段求出 高度，和单纯求每层节点的高度是一个道理。

 

题目五：判断一个树是否是二叉搜索树？

比较好理解的思路：中序遍历 以后的升序序列就是 二叉搜索树

bool isValidBST(TreeNode* root) {
    stack<TreeNode*>que;
    TreeNode * p = root;
    int pre=INT_MIN;
    int num=0;
    while (p|| que.empty()==false) {
        if(p){
            que.push(p);
            p=p->left;
        }
        else{
            TreeNode * node = que.top();
            que.pop();
            int val = node->val;
            printf("%d\n",val);
            if(num==1){
                if(val<=pre){
                    return false;
                }
            }
           
            num=1;
            pre=val;
            p=node->right;
        }
    }
    return true;
}

题目六：判断一个树是否为彻底二叉树？

思路：

能够借助队列

在两种状况下，必定不是彻底二叉树

状况1: 左子树为空，右子树不为空

状况2:开始叶子节点遍历的时候，若是不是叶子节点，就不是彻底二叉树

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;
            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 {
                leaf = true;
            }
        }
        return true;
    }

题目七：已知一棵彻底二叉树，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

思路1：若是时间复杂度为O(N)，那么就能够遍历这棵树，可是没有利用彻底二叉树的特色。

这里咱们每次看当前节点的右子树的是否达到了最底部。

1 若是达到了，那么当前点的左子树必定就是彻底二叉树，能够利用h^2-1求出左子树的节点数目，

而后递归求解右子树的节点个数。

2 若是没达到，那么当前节点的右子树必定是彻底二叉树，节点数为(h-1)^2-1 ,而后递归求解左子树的节点个数

3 递归终止条件为 到达叶节点，返回1

代码：

public class 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));
    }

    public static int bs(Node node, int l, int h) {
        if (l == h) {
            return 1;
        }
        if (mostLeftLevel(node.right, l + 1) == h) {
            return (1 << (h - l)) + bs(node.right, l + 1, h);
        } else {
            return (1 << (h - l - 1)) + bs(node.left, l + 1, h);
        }
    }

    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));

    }

}

 思路2:

一颗树为彻底二叉树就分为两种状况：

一种是左子树和右子树的最大高度同样，另外一种是左子树比右子树高度大1，因此能够根据这两种状况写递归：

第一种状况 左子树必定是彻底二叉树，第二种状况右子树必定是彻底二叉树：

int getDepth(TreeNode *node){
    int height = 0;
    while (node) {
        node = node->left;
        height++;
    }
    return height;
}

int getReturn(TreeNode * node){
    
    if(node==nullptr){
        return 0;
    }
    
    int h1= getDepth(node->left);
    int h2= getDepth(node->right);
    
    //若是 左右子树 最大高度相等，左边为彻底二叉树
    if(h1==h2){
       return (1<<h1)+ getReturn(node->right);
    }
    //右边为彻底二叉树
    else {
        return (1<<h2)+ getReturn(node->left);
    }
    
}

//彻底二叉树节点个数
int countNodes(TreeNode* root) {
    return getReturn(root);
}