README_EN.md

94. Binary Tree Inorder Traversal

中文文档

Description

Given a binary tree, return the inorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]

   1

    \

     2

    /

   3



Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?

Solutions

Python3

# 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 inorderTraversal(self, root: TreeNode) -> List[int]:
        def inorder(root):
            if root:
                inorder(root.left)
                res.append(root.val)
                inorder(root.right)
        res = []
        inorder(root)
        return res

Java

/**
 * 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> inorderTraversal(TreeNode root) {
        res = new ArrayList<>();
        inorder(root);
        return res;
    }

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

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