二叉树的递归,非递归前序/中序/后序操作

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package com.nowcoder.校招真题2016;    import java.util.Stack;    import java.util.HashMap;        public class BinTree {

//二叉树的三个属性;      private char date;   //节点值     private BinTree lchild;   //左子树     private BinTree rchild;     //右子树        public BinTree(char c) {            date = c;        }            // 先序遍历递归         public static void preOrder(BinTree t) {            if (t == null) {                return;            }            System.out.print(t.date);            preOrder(t.lchild);            preOrder(t.rchild);        }            // 中序遍历递归         public static void InOrder(BinTree t) {            if (t == null) {                return;            }            InOrder(t.lchild);            System.out.print(t.date);            InOrder(t.rchild);        }            // 后序遍历递归         public static void PostOrder(BinTree t) {            if (t == null) {                return;            }            PostOrder(t.lchild);            PostOrder(t.rchild);            System.out.print(t.date);        }            // 先序遍历非递归

            B             C

       D       E       F      G

    H    I   J   K   L   M   N   O

遍历顺序  ABDH ->I->E ->J->K->C->F->L->M->G->N->O

    public static void preOrder2(BinTree t) {

        Stack<BinTree> s = new Stack<BinTree>();            while (t != null || !s.empty()) {    //第一次运行时  t !=null    s.empty()成立            while (t != null) {                    System.out.print(t.date);                    s.push(t);   // 将T对象压入栈内                t = t.lchild;     将这个的左子树赋值给 T, 不停的循环并将左子树的节点压入到栈内,最后遍历到最后一个左子树            }                if (!s.empty()) {                    t = s.pop();  //                  t = t.rchild;                }            }        }

    // 中序遍历非递归         public static void InOrder2(BinTree t) {             Stack<BinTree> s = new Stack<BinTree>();           while (t != null || !s.empty()) {                while (t != null) {                    s.push(t);                    t = t.lchild;                }                if (!s.empty()) {                    t = s.pop();                    System.out.print(t.date);                    t = t.rchild;                }            }        }            // 后序遍历非递归         public static void PostOrder2(BinTree t) {            Stack<BinTree> s = new Stack<BinTree>();           Stack<Integer> s2 = new Stack<Integer>();            Integer i = new Integer(1);            while (t != null || !s.empty()) {                while (t != null) {                    s.push(t);                    s2.push(new Integer(0));                    t = t.lchild;                }                while (!s.empty() && s2.peek().equals(i)) {                    s2.pop();                    System.out.print(s.pop().date);                }                    if (!s.empty()) {                    s2.pop();                    s2.push(new Integer(1));                    t = s.peek();                    t = t.rchild;                }            }        }            public static void main(String[] args) {            BinTree b1 = new BinTree('a');            BinTree b2 = new BinTree('b');            BinTree b3 = new BinTree('c');            BinTree b4 = new BinTree('d');            BinTree b5 = new BinTree('e');                /**           *      a            *     / /           *    b   c           *   / /           *  d   e           */            b1.lchild = b2;            b1.rchild = b3;            b2.lchild = b4;            b2.rchild = b5;                BinTree.preOrder(b1);            System.out.println();            BinTree.preOrder2(b1);            System.out.println();            BinTree.InOrder(b1);            System.out.println();            BinTree.InOrder2(b1);            System.out.println();            BinTree.PostOrder(b1);            System.out.println();            BinTree.PostOrder2(b1);        }    }

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