算法通关村第七关|黄金挑战|迭代实现二叉树的前、中、后序遍历

1.迭代实现前序遍历

public List<Integer> preOrderTraversal(TreeNode root) {
	List<Integer> res = new ArrayList<Integer>();
	if (root == null) {
        return res;
    }
	Deque<TreeNode> stack = new LinkedList<TreeNode>();
	TreeNode node = root;
	while (!stack.isEmpty() || node != null) {
        while (node != null) {
            res.add(node.val);
            stack.push(node);
            node = node.left;
        }
        node = stack.pop();
        node = node.right;
    }
	return res;
}

2.迭代实现中序遍历

public List<Integer> inorderTraversal(TreeNode root) {
	List<Integer> res = new ArrayList<Integer>();
	Deque<TreeNode> stack = new LinkedList<TreeNode>();
	while (root != null || !stack.isEmpty()) {
        while (root != null) {
            stack.push(root);
            root = root.left;
        }
        root = stack.pop();
        res.add(root.val);
        root = root.right;
    }
	return res;
}

3.迭代实现后序遍历（反转法）

将后序遍历的结果反过来就是类似于前序遍历的结果，只不过前序遍历先找左孩子，后序反过来是先找右孩子。

public List<Integer> postOrderTraversal(TreeNode root) {
	List<Integer> res = new ArrayList<>();
	if (root == null) {
        return res;
    }
	Deque<TreeNode> stack = new LinkedList<TreeNode>();
	TreeNode node = root;
	while (!stack.isEmpty() || node != null) {
        while (node != null) {
            res.add(node.val);
            stack.push(node);
            node = node.right;
        }
        node = stack.pop();
        node = node.left;
    }
	Collections.reverse(res);
	return res;
}

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