迭代法实现二叉树的前，中，后序遍历

/**
 * @author wkn
 * @create 2021-12-03 9:55
 */
public class DepthFirstSearch {
    public static void main(String ... args){
        TreeNode root = new TreeNode(0);
        root.leftNode = new TreeNode(1);
        root.rightNode = new TreeNode(2);
        root.leftNode.rightNode = new TreeNode(4);
        List<Integer> preOrder = dfsPreOrder(root);
        System.out.println("先序遍历");
        show(preOrder);
        System.out.println("\n******************");
        System.out.println("中序遍历");
        List<Integer> inOrder = dfsInOrder(root);
        show(inOrder);
        System.out.println("\n******************");
        System.out.println("后序遍历");
        List<Integer> lastOrder = dfsLastOrder(root);
        show(lastOrder);

    }

    public static void show(List<Integer> list){
        for(int val : list){
            System.out.print(val+" ");
        }
    }

    public static List<Integer> dfsPreOrder(TreeNode root){
        if(root == null){
            return null;
        }
        List<Integer> result = new ArrayList<Integer>();
        Stack<TreeNode> treeNodes = new Stack<>();
        treeNodes.add(root);

        while(!treeNodes.isEmpty()){
            TreeNode pop = treeNodes.pop();
            result.add(pop.val);
            //后进先出，所以左子结点先出来
            if(pop.rightNode != null){
                treeNodes.push(pop.rightNode);
            }
            if(pop.leftNode != null){
                treeNodes.push(pop.leftNode);
            }
        }

        return result;
    }

    public static List<Integer> dfsInOrder(TreeNode root){
        if(root == null){
            return null;
        }
        List<Integer> result = new ArrayList<Integer>();
        Stack<TreeNode> treeNodes = new Stack<TreeNode>();

        while(root != null || !treeNodes.isEmpty()){
            while(root != null){
                treeNodes.push(root);
                root = root.leftNode;
            }
            root = treeNodes.pop();//每个结点自身又是中间的，又是父节点的子节点
            result.add(root.val);
            root = root.rightNode;//当前结点的右结点
        }

        return result;
    }

    public static List<Integer> dfsLastOrder(TreeNode root){
        if(root == null){
            return null;
        }
        List<Integer> result = new ArrayList<Integer>();
        Stack<TreeNode> treeNodes = new Stack<TreeNode>();
        TreeNode prev = null;

        while(root != null || !treeNodes.isEmpty()){
            while(root != null){
                treeNodes.push(root);
                root = root.leftNode;
            }

            root = treeNodes.pop();//第一次可能是最左结点，但是也是中间的结点
            if(root.rightNode == null || root.rightNode == prev){ //当前结点没有右节点 或者 当前结点的右节点已经添加到序列中了
                result.add(root.val);//可以将当前结点遍历进序列中
                prev = root;
                root = null;
            }else{  //当前结点还有右节点，将当前结点入栈，先去遍历右子树
                treeNodes.push(root);
                root = root.rightNode;
            }
        }
        return result;
    }

}

class TreeNode{
    int val;
    TreeNode leftNode;
    TreeNode rightNode;
    public TreeNode(){}
    public TreeNode(int val){
        this.val = val;
    }
    public TreeNode(int val, TreeNode leftNode, TreeNode rightNode){
        this.val = val;
        this.leftNode = leftNode;
        this.rightNode = rightNode;
    }
}

迭代的前序遍历也可以这样写：

    public static List<Integer> dfsPreOrder(TreeNode root){
        if(root == null){
            return null;
        }
        List<Integer> result = new ArrayList<Integer>();
        Stack<TreeNode> treeNodes = new Stack<>();

        /*while(!treeNodes.isEmpty()){
            TreeNode pop = treeNodes.pop();
            result.add(pop.val);
            //后进先出，所以左子结点先出来
            if(pop.rightNode != null){
                treeNodes.push(pop.rightNode);
            }
            if(pop.leftNode != null){
                treeNodes.push(pop.leftNode);
            }
        }*/

        while(root != null || !treeNodes.isEmpty()){
            while(root != null){
                result.add(root.val);
                treeNodes.push(root);
                root = root.leftNode;
            }

            root = treeNodes.pop();
            root = root.rightNode;
        }

        return result;
    }