144_二叉树的前序遍历

目录

144_二叉树的前序遍历

描述

给定一个二叉树,返回它的前序遍历。

示例:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [1,2,3]

进阶: 递归算法很简单,你可以通过迭代算法完成吗?

方法一:递归

Java 代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        preorderTraversal(root, res);
        return res;
    }
    
    private void preorderTraversal(TreeNode root, List<Integer> res) {
        if (root == null) {
            return;
        }
        
        res.add(root.val);
        preorderTraversal(root.left, res);
        preorderTraversal(root.right, res);
    }
}

复杂度分析:

  • 时间复杂度:\(O(n)\),其中,\(n\) 为二叉树节点的数目
  • 空间复杂度:\(O(n)\)

Python 代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        def dfs(root, ret):
            if root is None:
                return

            ret.append(root.val)
            dfs(root.left, ret)
            dfs(root.right, ret)

        ret = list()
        dfs(root, ret)
        return ret

复杂度分析同上。

方法二:非递归(使用栈)

Java 代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            
            if (cur.right != null) {
                stack.push(cur.right);
            }
            if (cur.left != null) {
                stack.push(cur.left);
            }
        }
        return res;
    }
}

复杂度分析:

  • 时间复杂度:\(O(n)\),其中,\(n\) 为二叉树节点的数目
  • 空间复杂度:\(O(h)\),其中,\(h\) 为二叉树的高度

Python 代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        
        ret, stack = [], [root]
        while len(stack) > 0:
            node = stack.pop()
            ret.append(node.val)
            
            if node.right is not None:
                stack.append(node.right)
            if node.left is not None:
                stack.append(node.left)
                
        return ret

复杂度分析同上。

posted @ 2018-10-11 23:36 StrongXGP 阅读(...) 评论(...) 编辑 收藏