day14 打卡二叉树的递归遍历 迭代遍历 统一迭代

144. 二叉树的前序遍历

144题目链接

1.递归遍历

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        preorder(root, list);
        return list;
    }

    public void preorder(TreeNode node, List<Integer> list) {
        if (node == null) return;
        list.add(node.val);
        preorder(node.left, list);
        preorder(node.right, list);
    }
}

2.迭代遍历

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if (root == null) return list;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = null;
        stack.push(root);
        while (!stack.isEmpty()) {
            cur = stack.pop();
            list.add(cur.val);
            if (cur.right != null) {
                stack.push(cur.right);
            }
            if (cur.left != null) {
                stack.push(cur.left);
            }
        }
        return list;
    }
}

3.统一迭代

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if (root == null) return list;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = null;
        stack.push(root);
        while (!stack.isEmpty()) {
            cur = stack.peek();
            if (cur != null) {
                // 弹出节点
                stack.pop();
                // 右
                if (cur.right != null) {
                    stack.push(cur.right);
                }
                // 左
                if (cur.left != null) {
                    stack.push(cur.left);
                }
                // 中
                stack.push(cur);
                // 标记
                stack.push(null);
            } else {
                // 弹出null
                stack.pop();
                // 弹出下一个，加入list
                list.add(stack.pop().val);
            } 
        }
        return list;
    }
}

145. 二叉树的后序遍历

145题目链接

1.递归遍历

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        postorder(root, list);
        return list;
    }

    public void postorder(TreeNode node, List<Integer> list) {
        if (node == null) return;
        postorder(node.left, list);
        postorder(node.right, list);
        list.add(node.val);
    }
}

2.迭代遍历

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if (root == null) return list;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = null;
        stack.push(root);
        while (!stack.isEmpty()) {
            cur = stack.pop();
            list.add(cur.val);
            if (cur.left != null) {
                stack.push(cur.left);
            }
            if (cur.right != null) {
                stack.push(cur.right);
            }
        }
        Collections.reverse(list);
        return list;
    }
}

3.统一迭代

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if (root == null) return list;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = null;
        stack.push(root);
        while (!stack.isEmpty()) {
            cur = stack.peek();
            if (cur != null) {
                // 弹出节点
                stack.pop();
                // 中
                stack.push(cur);
                stack.push(null);
                // 右
                if (cur.right != null) stack.push(cur.right);
                // 左
                if (cur.left != null) stack.push(cur.left);
            } else {
                // 弹出null
                stack.pop();
                // 弹出下一个，加入list
                list.add(stack.pop().val);
            }
        }
        return list;
    }
}

94. 二叉树的中序遍历

94题目链接
1.递归遍历

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        inorder(root, list);
        return list;
    }

    public void inorder(TreeNode node, List<Integer> list) {
        if (node == null) return;
        inorder(node.left, list);
        list.add(node.val);
        inorder(node.right, list);
    }
}

2.迭代遍历

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while (cur != null || !stack.isEmpty()) {
            if (cur == null) {
                cur = stack.pop();
                list.add(cur.val);
                cur = cur.right;
            } else {
                stack.push(cur);
                cur = cur.left;
            }
        }
        return list;
    }
}

3.统一迭代

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if (root == null) return list;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = null;
        stack.push(root);
        while (!stack.isEmpty()) {
            cur = stack.peek();
            if (cur == null) {
                // 弹出null
                stack.pop();
                // 弹出下一个，加入list
                list.add(stack.pop().val);
            } else {
                // 弹出节点
                stack.pop();
                // 右
                if (cur.right != null) stack.push(cur.right);
                // 中
                stack.push(cur);
                stack.push(null);
                // 左
                if (cur.left != null) stack.push(cur.left);
            }
        }
        return list;
    }
}

参考资料

代码随想录