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题目链接：leetcode.com/problems/bi…

1. 题目介绍（Binary Tree Inorder Traversal）

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

【Translate】： 给定二叉树的根，返回其节点值的中序遍历。

【测试用例】：

【条件约束】：

【跟踪】：

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

【Translate】： 递归求解很简单，你能迭代求解吗

2. 题解

这个题就没什么好讲的了， 属于数据结构的基础，即 左-root-右.

2.1 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> inorderTraversal(TreeNode root){
        List<Integer> res = new ArrayList<>();
        inorder(root, res);
        return res;
    }
    /**
     * 递归三步骤、
     * 1. 确定递归函数的参数和返回值；
     * 2. 确定终止条件；
     * 3. 确定单层递归的逻辑
     */
    private void inorder(TreeNode root, List<Integer> res) {
        if(root == null){
            return;
        }
        inorder(root.left, res);
        res.add(root.val);
        inorder(root.right, res);

    }

}

or

/**
 * 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 {
    List<Integer> list = new ArrayList<>();
    
    public List<Integer> inorderTraversal(TreeNode root) {
        if (root != null){
            inorderTraversal(root.left);
            list.add(root.val);
            inorderTraversal(root.right);
        }
        return list;
    }
}

2.2 栈

程序解释流程图来自于 YaoFrankie 在 Iterative solution in Java - simple and readable 中的comment.

/**
 * 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> inorderTraversal(TreeNode root) {
        
        TreeNode node = root;
        List<Integer> list = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        
        
        while (node != null || !stack.empty()){
            while (node != null){
                stack.push(node);
                node = node.left;
            }
            
            node = stack.pop();
            list.add(node.val);
            node = node.right;
        }
        
        return list;
    }
}

3. 参考资料

[1] 前序遍历 (preorder traversal) - 中序遍历 (inorder traversal) - 后序遍历 (postorder traversal) | CSDN [2] Java实现二叉树的递归遍历(前序遍历、中序遍历、后序遍历) | CSDN