遍历二叉树的模板方法

二叉树

示意图

遍历序列

• 先序遍历序列：[4、2、1、3、6、5、7]
• 中序遍历序列：[1、2、3、4、5、6、7]
• 后续遍历序列：[1、3、2、5、7、6、4]

遍历方案

二叉树节点

class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;

	TreeNode(int val) {
		this.val = val;
	}
}

先序遍历（深度优先）

• 递归方案

public static void preorder(TreeNode root, List<TreeNode> result) {

	if (root == null) {
		return;
	}

	result.add(root);
	preorder(root.left, result);
	preorder(root.right, result);
}

• 非递归方案

public void preorder(TreeNode root, List<TreeNode> result) {

	TreeNode cur = root;
	Stack<TreeNode> stack = new Stack<>();

	while (cur != null || !stack.isEmpty()) {

		if (cur != null) {
			result.add(cur);

			stack.push(cur);
			cur = cur.left;
		}
		else {
			cur = stack.pop();
			cur = cur.right;
		}
	}
}

中序遍历

• 递归方案

public static void inorder(TreeNode root, List<TreeNode> result) {

	if (root == null) {
		return;
	}

	inorder(root.left, result);
	result.add(root);
	inorder(root.right, result);
}

• 非递归方案

public static void inorder(TreeNode root, List<TreeNode> result) {

	TreeNode cur = root;
	Stack<TreeNode> stack = new Stack<>();

	while (cur != null || !stack.isEmpty()) {
		if (cur != null) {
			stack.push(cur);
			cur = cur.left;
		}
		else {
			cur = stack.pop();
			result.add(cur);
			cur = cur.right;
		}
	}
}

后续遍历

• 递归方案

public static void postorder(TreeNode root, List<TreeNode> result) {

	if (root == null) {
		return;
	}

	postorder(root.left, result);
	postorder(root.right, result);
	result.add(root);
}

• 非递归方案

public static void postorder(TreeNode root, List<TreeNode> result) {

	TreeNode cur = root, last = null;
	Stack<TreeNode> stack = new Stack<>();

	while (cur != null || !stack.isEmpty()) {
		if (cur != null) {
			stack.push(cur);
			cur = cur.left;
		}
		else {
			cur = stack.peek();
			if (cur.right == null || cur.right == last) {
				result.add(cur);

				stack.pop();
				last = cur;
				cur = null;
			}
			else {
				cur = cur.right;
			}
		}
	}
}

层次遍历（广度优先遍历）

public List<TreeNode> levelOrder(TreeNode root) {

	List<TreeNode> list = new LinkedList<>();

	if (root == null) {
		return list;
	}

	Queue<TreeNode> queue = new LinkedList<>();
	queue.offer(root);

	while (!queue.isEmpty()) {

		TreeNode node = queue.poll();
		list.add(node);

		if (node.left != null) {
			queue.offer(node.left);
		}

		if (node.right != null) {
			queue.offer(node.right);
		}
	}

	return list;
}