构造树如下:
其中二叉树节点类
/** 二叉树节点 */public class BTNode { private char key; private BTNode left, right; public BTNode(char key) { this(key, null, null); } public BTNode(char key, BTNode left, BTNode right) { this.key = key; this.left = left; this.right = right; } public char getKey() { return key; } public void setKey(char key) { this.key = key; } public BTNode getLeft() { return left; } public void setLeft(BTNode left) { this.left = left; } public BTNode getRight() { return right; } public void setRight(BTNode right) { this.right = right; }}
遍历二叉树
/** 二叉树遍历 */public class BinTree { protected BTNode root; public BinTree(BTNode root) { this.root = root; } public BTNode getRoot() { return root; } /** 构造树 */ public static BTNode init() { BTNode a = new BTNode('A'); BTNode b = new BTNode('B', null, a); BTNode c = new BTNode('C'); BTNode d = new BTNode('D', b, c); BTNode e = new BTNode('E'); BTNode f = new BTNode('F', e, null); BTNode g = new BTNode('G', null, f); BTNode h = new BTNode('H', d, g); return h;// root } /** 访问节点 */ public static void visit(BTNode p) { System.out.print(p.getKey() + " "); } /** 递归实现前序遍历 */ protected static void preorder(BTNode p) { if (p != null) { visit(p); preorder(p.getLeft()); preorder(p.getRight()); } } /** 递归实现中序遍历 */ protected static void inorder(BTNode p) { if (p != null) { inorder(p.getLeft()); visit(p); inorder(p.getRight()); } } /** 递归实现后序遍历 */ protected static void posTorder(BTNode p) { if (p != null) { posTorder(p.getLeft()); posTorder(p.getRight()); visit(p); } } /** 非递归实现前序遍历 */ protected static void iterativePreorder(BTNode p) { Stack stack = new Stack(); if (p != null) { stack.push(p); while (!stack.empty()) { p = stack.pop(); visit(p); if (p.getRight() != null) stack.push(p.getRight()); if (p.getLeft() != null) stack.push(p.getLeft()); } } } /** 非递归实现后序遍历 */ protected static void iterativePosTorder(BTNode p) { BTNode q = p; Stack stack = new Stack(); while (p != null) { // 左子树入栈 for (; p.getLeft() != null; p = p.getLeft()) stack.push(p); // 当前节点无右子或右子已经输出 while (p != null && (p.getRight() == null || p.getRight() == q)) { visit(p); q = p;// 记录上一个已输出节点 if (stack.empty()) return; p = stack.pop(); } // 处理右子 stack.push(p); p = p.getRight(); } } /** 非递归实现中序遍历 */ protected static void iterativeInorder(BTNode p) { Stack stack = new Stack(); while (p != null) { while (p != null) { if (p.getRight() != null) stack.push(p.getRight());// 当前节点右子入栈 你会发现，曾经以为很难做到的事情，
