二叉树的递归,非递归前序/中序/后序操作

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本文详细介绍了二叉树的遍历方法，包括先序、中序和后序遍历的递归与非递归实现，并通过具体示例展示了遍历过程。

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);    
    }    
}