二叉树三种遍历（递归以及非递归实现）

package com.shiyeqiang.tree;import java.util.Stack;public class BiTree {public static void main(String[] args) {// 首先构造叶子节点BiTree leafA1 = new BiTree(4);BiTree leafA2 = new BiTree(5);BiTree leafB1 = new BiTree(6);BiTree leafB2 = new BiTree(7);// 构建二叉树的结构BiTree treeA = new BiTree(2, leafA1, leafA2);BiTree treeB = new BiTree(3, leafB1, leafB2);BiTree tree = new BiTree(1, treeA, treeB);System.out.println("递归前序遍历二叉树结果: ");preOrder(tree);System.out.println();System.out.println("非递归前序遍历二叉树结果: ");iterativePreOrder(tree);System.out.println();System.out.println("非递归中序遍历二叉树结果: ");iterativeInOrder(tree);System.out.println();System.out.println("递归兴许遍历二叉树结果: ");iterativePostOrder(tree);}private BiTree leftTree;private BiTree rightTree;private Object data;public BiTree() {}public BiTree(Object data) {this.data = data;}public BiTree(Object data, BiTree leftTree, BiTree rightTree) {this.data = data;this.leftTree = leftTree;this.rightTree = rightTree;}public static void visit(Object data) { // 訪问二叉树的数据System.out.print(data + "  ");}public static void preOrder(BiTree tree) {// 递归实现前序遍历二叉树if (tree != null) {visit(tree.data);if (tree.leftTree != null)preOrder(tree.leftTree);if (tree.rightTree != null)preOrder(tree.rightTree);}}public static void inOrder(BiTree tree) {// 递归实现中序遍历二叉树(左-根--右边)if (tree != null) {if (tree.leftTree != null)inOrder(tree.leftTree);visit(tree.data);if (tree.rightTree != null)inOrder(tree.rightTree);}}public static void postOrder(BiTree tree) {// 递归实现后序遍历二叉树if (tree != null) {if (tree.leftTree != null)postOrder(tree.leftTree);if (tree.rightTree != null)postOrder(tree.rightTree);visit(tree.data);}}// 非递归实现前序遍历，根----左子树---右子树public static void iterativePreOrder(BiTree tree) {Stack<BiTree> stack = new Stack<BiTree>();if (tree != null) {stack.push(tree);while (!stack.isEmpty()) {tree = stack.pop();visit(tree.data);if (tree.rightTree != null)stack.push(tree.rightTree); // 首先必须是右子树入栈，if (tree.leftTree != null)stack.push(tree.leftTree);}}}// 非递归实现中序遍历，左---根---右public static void iterativeInOrder(BiTree tree) {Stack<BiTree> stack = new Stack<BiTree>();// 主要是依据入栈以及出栈的思想实现的//以一个1,2,3,4,5,6,7的满二叉树为例：中序遍历例如以下：//右子树3入栈，根节点1入栈，右子树5入栈，根节点2入栈，根节点4入栈。//左子树4出栈，根节点2出栈，5出战，依照上述规则又一次遍历右子树5while (tree != null) {while (tree != null) {if (tree.rightTree != null)stack.push(tree.rightTree);// 当前节点右子入栈stack.push(tree);// 当前节点入栈tree = tree.leftTree;}tree = stack.pop();// 以下为訪问的是叶子节点while (!stack.empty() && tree.rightTree == null) {visit(tree.data);tree = stack.pop();}visit(tree.data);if (!stack.empty()) {tree = stack.pop(); // 下一次遍历,事实上在这个时候遍历的是右子树把} else {tree = null;}}}// 非递归实现后序遍历public static void iterativePostOrder(BiTree tree) {BiTree tempTree = tree;Stack<BiTree> stack = new Stack<BiTree>();while (tree != null) {// 左子树入栈for (; tree.leftTree != null; tree = tree.leftTree)stack.push(tree);// 当前节点无右子或右子已经输出while (tree != null&& (tree.rightTree == null || tree.rightTree == tempTree)) {visit(tree.data);tempTree = tree;// 记录上一个已输出节点if (stack.empty())return;tree = stack.pop(); //一般处理根节点的时候均是先出战 ，然后再入栈}// 处理右子stack.push(tree);tree = tree.rightTree;}}
}