Java语言实现非递归实现树的前中后序遍历总结

public class Node {
  int val;
  Node left;
  Node right;
  Node parent;
  Node() {}
  Node(int val) {
    this.val = val;
  }
}

public void preOrder(Node head) {
    System.out.println("pre-order:");
    if(head != null) {
      Stack<Node> stack = new Stack<>();
      stack.push(head);
      while(!stack.isEmpty()) {
        head = stack.pop();
        System.out.print(head.val + " ");
        if (head.right != null)
          stack.push(head.right);
        if (head.left != null)
          stack.push(head.left);
      }
    }
  }

public static void inOrderTraverse(Node head) {
    System.out.println("in-order:");
    if(head != null) {
      Stack<Node> stack = new Stack<>();
      while(!stack.isEmpty() || head != null) {
        if(head != null) {
          // 当前节点不为空, 将自己压进栈并将自己的左孩子作为当前节点（压入左边界）
          stack.push(head);
          head = head.left;
        }else {
          // 当前节点为空（没有左孩子了）, 将栈顶元素弹出作为当前节点, 并将当前节点的右孩子压进栈
          head = stack.pop();
          System.out.print(head.val + " ");
          head = head.right;
        }
      }
    }
  }

public static void posOrderTraverse(Node head) {
    System.out.println("pos-order");
    if(head != null) {
      Stack<Node> stack1 = new Stack<>();
      Stack<Node> stack2 = new Stack<>();   // 辅助栈，存储 根 -> 右 -> 左 的结果
      stack1.push(head);
      while(!stack1.isEmpty()) {
        head = stack1.pop();
        stack2.push(head);
        // 有左孩子就先压入左孩子
        if(head.left != null)
          stack1.push(head.left);
        // 有右孩子就后压入右孩子
        if(head.right != null)
          stack1.push(head.right);
      }
      // 逆序打印 根 -> 右 -> 左 的结果，就是后序遍历的结果
      while(!stack2.isEmpty())
        System.out.print(stack2.pop().val + " ");
    }
  }