import java.util.ArrayDeque;import java.util.Queue;import java.util.Stack;/** * Created by renshuang on 10/11/13. */public class TestTree { /** * @param args */ public static void main(String args[]) { TestTree tree = new TestTree(); tree.add(11); tree.add(8); tree.add(14); tree.add(6); tree.add(9); tree.add(15); tree.add(18); tree.add(10); tree.add(8); int arr[] = {5, 7, 6, 9, 11, 10, 8}; // Yes // int arr[] = {7, 4, 6, 5}; // Not System.out.println(verfyBst(arr, 0, arr.length)); tree.prePrint(tree.root); // tree.midPrint(tree.root); } public void breadFirst() { Queue<Node> queue = new ArrayDeque<Node>(); if (root != null) { queue.add(root); } while (!queue.isEmpty()) { Node current = queue.poll(); System.out.println(current.data); if (current.left != null) { queue.add(current.left); } if (current.right != null) { queue.add(current.right); } } } private Node root; /** * 11 * / \ * 8 14 * / \ / \ * 6 9 15 * / \ \ * 8 10 18 * * @param data */ public void add(int data) { if (root == null) { root = new Node(data, null, null); } else { Node current = root; Node parent = null; while (current != null) { parent = current; if (current.data > data) { current = current.left; } else { current = current.right; } } if (parent != null) { if (parent.data > data) { parent.left = new Node(data, null, null); } else { parent.right = new Node(data, null, null); } } } } /** * 前序遍历 跟广度优先遍历唯一不同的是这个用到了stack * * @param root */ public void prePrint(Node root) { Stack<Node> stack = new Stack<Node>(); if (root != null) { stack.add(root); } while (!stack.isEmpty()) { Node current = stack.pop(); System.out.println(current.data); if (current.right != null) { stack.push(current.right); } if (current.left != null) { stack.push(current.left); } } } /** * 算法思想：三种算法的思想都是让root的Left的Left的Left全都入栈。所以第一个while循环的逻辑，都是相同的。 * 下面详细分析第2个while循环，这是一个出栈动作，只要栈不为空，就始终要弹出栈顶元素， * 由于我们之前入栈的都是Left节点，所以每次在出栈的时候，我们都要考虑Right节点是否存在。 * 因为前序/后序/中序遍历顺序的不同，所以在具体的实现上有略为区别。 * * @param root */ public void midPrint(Node root) { Stack<Node> stack = new Stack<Node>(); Node current = root; /** * 把当前所有左子树入栈 */ while (current != null) { stack.push(current); current = current.left; } while (stack.size() > 0) { /** * 出站打印当前内容 */ current = stack.pop(); System.out.println(current.data); /** * 如果当前节点有右子树则把当前右孩子如栈， * 再判断当前右孩子有没有左孩子 */ if (current.right != null) { stack.push(current.right); /* 必须下移右孩否泽 出问题，例如节点9 */ current = current.right; while (current != null) { if (current.left != null) { stack.push(current.left); } current = current.left; } } } } public void midTest(Node root) { Stack<Node> stack = new Stack<Node>(); Node current = root; while (current != null) { stack.add(current); current = current.left; } while (stack.size() > 0) { current = stack.pop(); System.out.println(current.data); if (current.right != null) { stack.push(current.right); current = current.right; while (current != null) { if (current.left != null) { stack.push(current.left); } current = current.left; } } } } public static boolean verfyBst(int[] data, int start, int end) { if (data == null || start > end) { return false; } if (start == end) { return true; } int root = data[end - 1]; int i = start; for (; i < end - 1; i++) { if (data[i] >= root) { break; } } int j = i; for (; j < end - 1; j++) { if (data[j] < root) { return false; } } return verfyBst(data, start, i) && verfyBst(data, i, end - 1); } class Node { int data; Node left; Node right; public Node() { } public Node(int data, Node left, Node right) { this.data = data; this.right = right; this.left = left; } }}
