94. 二叉树的中序遍历

https://leetcode-cn.com/problems/binary-tree-inorder-traversal/

Java

/*
 * @Author: Goog Tech
 * @Date: 2020-07-26 21:00:03
 * @Description: https://leetcode-cn.com/problems/binary-tree-inorder-traversal/comments/
 * @FilePath: \leetcode-googtech\#94. Binary Tree Inorder Traversal\Solution.java
 */

/*
 * @lc app=leetcode.cn id=94 lang=java
 *
 * [94] 二叉树的中序遍历
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

class Solution {

    // 递归法
    List<Integer> list = new ArrayList<>();
    public List<Integer> inorderTraversal(TreeNode root) {
        if(root != null) {
            inorderTraversal(root.left);
            list.add(root.val);
            inorderTraversal(root.right);
        }
        // [] + [1] + [3,2]
        return list;
    }

    // 迭代法
    public List<Integer> inorderTraversal(TreeNode root) {
        // 初始化辅助栈及结果列表
        Stack<TreeNode> stack = new Stack<>();
        List<Integer> result = new LinkedList<>();
        // 判断头节点是否为空
        if(root == null) return result;
        // 遍历辅助栈
        while(root != null || !stack.isEmpty()) {
            // 遍历根节点的左孩子节点并将其依次压入栈中,直到为空然后进入操作2
            if(root != null) { // 操作1
                stack.push(root);
                root = root.left;
            // 弹出栈顶元素,若其有右孩子,则将右孩子节点压入栈中,随后重复操作1
            }else { // 操作2
                root = stack.pop();
                result.add(root.val);
                root = root.right;
            }
        }
        return result;
    }
}

// @lc code=end

Python

'''
@Author: Goog Tech
@Date: 2020-07-26 19:19:05
@Description: https://leetcode-cn.com/problems/binary-tree-inorder-traversal/
@FilePath: \leetcode-googtech\#94. Binary Tree Inorder Traversal\94.二叉树的中序遍历.py
'''

#
# @lc app=leetcode.cn id=94 lang=python
#
# [94] 二叉树的中序遍历
#

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

class Solution(object):

    # 递归法
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None: return []
        # [] + [1] + [2,3]
        return self.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right)

    # 迭代法
    def inorderTraversal(self, root):
        # 初始化辅助栈及结果列表
        result, stack = [], []
        # 判断头节点是否为空
        if root is None: return result
        currentNode = root
        # 遍历辅助栈
        while stack or currentNode:
            # 遍历根节点的左孩子节点并将其依次压入栈中,直到为空然后进入操作2
            if currentNode is not None: # 操作1
                stack.append(currentNode)
                currentNode = currentNode.left
            # 弹出栈顶元素,若其有右孩子,则将右孩子节点压入栈中,随后重复操作1
            else: # 操作2
                currentNode = stack.pop()
                result.append(currentNode.val)
                currentNode = currentNode.right
        return result

# @lc code=end