采用递归方式实现

节点类

public class Node {private int value;//父节点private Node fNode;//左节点private Node left;//右节点private Node right;//是否已经打印过private boolean sign = false;public Node() {}public boolean isSign() {return sign;}public void setSign(boolean sign) {this.sign = sign;}public Node getfNode() {return fNode;}public void setfNode(Node fNode) {this.fNode = fNode;}public Node getLeft() {return left;}public void setLeft(Node left) {this.left = left;//设置左节点的父类left.setfNode(this);}public Node getRight() {return right;}public void setRight(Node right) {this.right = right;//设置右节点的父类right.setfNode(this);}public int getValue() {return value;}public Node setValue(int value) {this.value = value;return this;}
}

递归方法 

//后序遍历二叉树
public static void reducer(Node node) {if (node == null) return;//判断当前节点是否已经打印，如果没有就进去判断它的左右节点if (!node.isSign()) {/*if:如果左节点不为空并且没有被打印，那么就说明该节点未被调用过，进行递归操作else if: 判断完左节点就判断右节点，这是后续遍历的操作顺序else : 如果左节点和右节点都打印了，那么就打印当前节点的值，并且回退到上一个节点（父节点）*/if (node.getLeft() != null && !node.getLeft().isSign()) {reducer(node.getLeft());} else if (node.getRight() != null && !node.getRight().isSign()) {reducer(node.getRight());} else {System.out.print(node.getValue());node.setSign(true);//如果父节点为空说明已经遍历完了if (node.getfNode() != null) {reducer(node.getfNode());} else {return;}}}
}

测试 

public class BinaryIterator {public static void main(String[] args) {//构建顶级节点的左节点Node l = new Node().setValue(2);l.setLeft(new Node().setValue(3));l.setRight(new Node().setValue(4));//构建顶级节点的右节点Node r = new Node().setValue(5);r.setLeft(new Node().setValue(6));r.setRight(new Node().setValue(7));//顶节点（起始节点）Node node = new Node();node.setLeft(l);node.setRight(r);node.setValue(1);//测试reducer(node);//rest:3426751}}

草图（想要透彻理解，必须要慢慢根据图进行推演）