二叉树前中、后、序遍历的非递归遍历

package frank.test3;

import java.util.Stack;
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int val) { this.val = val; }
}
public class test {
    //前序
    public static void preOrder(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while (!stack.isEmpty() || cur != null) {
            while (cur != null) {
                System.out.println(cur.val);
                stack.push(cur);
                cur = cur.left;
            }
            TreeNode top = stack.pop();
            cur = top.right;
        }
    }
    //中序
    public static void inOrder(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while (!stack.isEmpty() || cur != null) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            TreeNode top = stack.pop();
            System.out.println(top.val);
            cur = top.right;
        }
    }
    //后序
    public static void postOrder(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        TreeNode pre = cur.left;//上一个节点
        while(!stack.isEmpty() || cur != null) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            TreeNode top = stack.peek();
            if (top.right == null || top.right == pre) {
                //从左子树回来 但由于右子树为空，可以看作从右子树回来 叶子节点
                //或者从右子树回来
                stack.pop();
                pre = top;
                System.out.println(top.val);
            } else  {
                //从左子树回来
                cur = top.right;
            }
        }
    }

    public static void main(String[] args) {
        TreeNode a = new TreeNode(1);
        TreeNode b = new TreeNode(2);
        TreeNode c = new TreeNode(3);
        TreeNode d = new TreeNode(4);
        TreeNode e = new TreeNode(5);
        a.left = b;a.right = c;
        b.left = d;b.right = e;
        postOrder(a);
    }
}