二叉树基本算法（上）
1、链表附加题
给定两个可能有环也可能无环的单链表，头节点head1和head2。请实现一个函数，如果两个链表相交，请返回相交的 第一个节点。如果不相交，返回null
【要求】如果两个链表长度之和为N，时间复杂度请达到O(N)，额外空间复杂度 请达到O(1)。
package class10;public class Code01_FindFirstIntersectNode {public static class Node {public int value;public Node next;public Node(int data) {this.value = data;}}public static Node getIntersectNode(Node head1, Node head2) {if (head1 == null || head2 == null) {return null;}Node loop1 = getLoopNode(head1);Node loop2 = getLoopNode(head2);if (loop1 == null && loop2 == null) {return noLoop(head1, head2);}if (loop1 != null && loop2 != null) {return bothLoop(head1, loop1, head2, loop2);}return null;}// 找到链表第一个入环节点，如果无环，返回nullpublic static Node getLoopNode(Node head) {if (head == null || head.next == null || head.next.next == null) {return null;}// n1 慢 n2 快Node slow = head.next; // n1 -> slowNode fast = head.next.next; // n2 -> fastwhile (slow != fast) {if (fast.next == null || fast.next.next == null) {return null;}fast = fast.next.next;slow = slow.next;}// slow fast 相遇fast = head; // n2 -> walk again from headwhile (slow != fast) {slow = slow.next;fast = fast.next;}return slow;}// 如果两个链表都无环，返回第一个相交节点，如果不想交，返回nullpublic static Node noLoop(Node head1, Node head2) {if (head1 == null || head2 == null) {return null;}Node cur1 = head1;Node cur2 = head2;int n = 0;while (cur1.next != null) {n++;cur1 = cur1.next;}while (cur2.next != null) {n--;cur2 = cur2.next;}if (cur1 != cur2) {return null;}// n : 链表1长度减去链表2长度的值cur1 = n > 0 ? head1 : head2; // 谁长，谁的头变成cur1cur2 = cur1 == head1 ? head2 : head1; // 谁短，谁的头变成cur2n = Math.abs(n);while (n != 0) {n--;cur1 = cur1.next;}while (cur1 != cur2) {cur1 = cur1.next;cur2 = cur2.next;}return cur1;}// 两个有环链表，返回第一个相交节点，如果不想交返回nullpublic static Node bothLoop(Node head1, Node loop1, Node head2, Node loop2) {Node cur1 = null;Node cur2 = null;if (loop1 == loop2) {cur1 = head1;cur2 = head2;int n = 0;while (cur1 != loop1) {n++;cur1 = cur1.next;}while (cur2 != loop2) {n--;cur2 = cur2.next;}cur1 = n > 0 ? head1 : head2;cur2 = cur1 == head1 ? head2 : head1;n = Math.abs(n);while (n != 0) {n--;cur1 = cur1.next;}while (cur1 != cur2) {cur1 = cur1.next;cur2 = cur2.next;}return cur1;} else {cur1 = loop1.next;while (cur1 != loop1) {if (cur1 == loop2) {return loop1;}cur1 = cur1.next;}return null;}}public static void main(String[] args) {// 1->2->3->4->5->6->7->nullNode head1 = new Node(1);head1.next = new Node(2);head1.next.next = new Node(3);head1.next.next.next = new Node(4);head1.next.next.next.next = new Node(5);head1.next.next.next.next.next = new Node(6);head1.next.next.next.next.next.next = new Node(7);// 0->9->8->6->7->nullNode head2 = new Node(0);head2.next = new Node(9);head2.next.next = new Node(8);head2.next.next.next = head1.next.next.next.next.next; // 8->6System.out.println(getIntersectNode(head1, head2).value);// 1->2->3->4->5->6->7->4...head1 = new Node(1);head1.next = new Node(2);head1.next.next = new Node(3);head1.next.next.next = new Node(4);head1.next.next.next.next = new Node(5);head1.next.next.next.next.next = new Node(6);head1.next.next.next.next.next.next = new Node(7);head1.next.next.next.next.next.next = head1.next.next.next; // 7->4// 0->9->8->2...head2 = new Node(0);head2.next = new Node(9);head2.next.next = new Node(8);head2.next.next.next = head1.next; // 8->2System.out.println(getIntersectNode(head1, head2).value);// 0->9->8->6->4->5->6..head2 = new Node(0);head2.next = new Node(9);head2.next.next = new Node(8);head2.next.next.next = head1.next.next.next.next.next; // 8->6System.out.println(getIntersectNode(head1, head2).value);}}
2、递归遍历二叉树
package class10;public class Code02_RecursiveTraversalBT {public static class Node {public int value;public Node left;public Node right;public Node(int v) {value = v;}}public static void f(Node head) {if (head == null) {return;}// 1f(head.left);// 2f(head.right);// 3}// 先序打印所有节点public static void pre(Node head) {if (head == null) {return;}System.out.println(head.value);pre(head.left);pre(head.right);}public static void in(Node head) {if (head == null) {return;}in(head.left);System.out.println(head.value);in(head.right);}public static void pos(Node head) {if (head == null) {return;}pos(head.left);pos(head.right);System.out.println(head.value);}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);pre(head);System.out.println("========");in(head);System.out.println("========");pos(head);System.out.println("========");}}
3、非递归遍历二叉树
先序遍历：
（1）头结点入栈。
（2）在一个循环中，栈顶元素出栈，记为cur，输出该节点。
（3）cur存在右孩子则右孩子入栈，存在左孩子则左孩子入栈，再进行下一次循环。
public static void pre(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();}
后序遍历：
先序遍历的出栈顺序为：头左右，而头右左的逆置为：左右头，即为后序遍历顺序。因此，可将先序遍历稍作修改，然后逆置即可得到后序遍历。
（1）设置两个栈，头结点入A栈。
（2）在一个循环中，A栈栈顶元素出栈，再入B栈。
（3）存在左孩子则左孩子入A栈，存在右孩子则右孩子入A栈，再执行下一次循环。
（4）B栈元素全部出栈。
public static void pos(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(); // 头 右 左s2.push(head);if (head.left != null) {s1.push(head.left);}if (head.right != null) {s1.push(head.right);}}// 左 右 头while (!s2.isEmpty()) {System.out.print(s2.pop().value + " ");}}System.out.println();}
中序遍历：
（1）设二叉树当前节点为cur，cur的整条左边界入栈。
（2）栈依次弹出节点，并打印，若弹出的该节点存在右孩子，则右孩子设为cur重复（1），若不存在右孩子，则栈继续弹出元素。
public static void in(Node cur) {System.out.print("in-order: ");if (cur != null) {Stack<Node> stack = new Stack <Node>();while (!stack.isEmpty() || cur != null) {if (cur != null) {stack.push(cur);cur = cur.left;} else {cur = stack.pop();System.out.print(cur.value + " ");cur = cur.right;}}}System.out.println();}