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144. 二叉树的前序遍历

English Version

题目描述

给你二叉树的根节点 root ，返回它节点值的 前序 遍历。

 

示例 1：

输入：root = [1,null,2,3]
输出：[1,2,3]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

示例 4：

输入：root = [1,2]
输出：[1,2]

示例 5：

输入：root = [1,null,2]
输出：[1,2]

 

提示：

• 树中节点数目在范围 [0, 100] 内
• -100 <= Node.val <= 100

 

进阶：递归算法很简单，你可以通过迭代算法完成吗？

解法

递归遍历或利用栈实现非递归遍历。

非递归的思路如下：

1. 定义一个栈，先将根节点压入栈
2. 若栈不为空，每次从栈中弹出一个节点
3. 处理该节点
4. 先把节点右孩子压入栈，接着把节点左孩子压入栈（如果有孩子节点）
5. 重复 2-4
6. 返回结果

Python3

递归：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        def preorder(root):
            if root:
                res.append(root.val)
                preorder(root.left)
                preorder(root.right)
        res = []
        preorder(root)
        return res

非递归：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        if root is None:
            return []
        res = []
        s = [root]
        while s:
            node = s.pop()
            res.append(node.val)
            if node.right:
                s.append(node.right)
            if node.left:
                s.append(node.left)
        return res

Java

递归：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {

    private List<Integer> res;

    public List<Integer> preorderTraversal(TreeNode root) {
        res = new ArrayList<>();
        preorder(root);
        return res;
    }

    private void preorder(TreeNode root) {
        if (root != null) {
            res.add(root.val);
            preorder(root.left);
            preorder(root.right);
        }
    }
}

非递归：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        if (root == null) {
            return Collections.emptyList();
        }
        List<Integer> res = new ArrayList<>();
        Deque<TreeNode> s = new ArrayDeque<>();
        s.push(root);
        while (!s.isEmpty()) {
            TreeNode node = s.pop();
            res.add(node.val);
            if (node.right != null) {
                s.push(node.right);
            }
            if (node.left != null) {
                s.push(node.left);
            }
        }
        return res;
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        if (root == nullptr) return res;
        stack<TreeNode*> s;
        s.push(root);
        while (!s.empty()) {
            TreeNode *node = s.top();
            s.pop();
            res.push_back(node->val);
            if (node->right) s.push(node->right);
            if (node->left) s.push(node->left);
        }
        return res;
    }
};

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