LeetCode 94. Binary Tree Inorder Traversal

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

For example:
Given binary tree [1,null,2,3],
1
\
2
/
3
return [1,3,2].

分析：中序遍历，先遍历左子树，再输出中间，再遍历右子树～

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> result;
    vector<int> inorderTraversal(TreeNode* root) {
        dfs(root);
        return result;
    }
    void dfs(TreeNode* root) {
        if(root == NULL) return ;
        dfs(root->left);
        result.push_back(root->val);
        dfs(root->right);
    }
};

/**

* Definition for a binary tree node.

* struct TreeNode {

* int val;

* TreeNode *left;

* TreeNode *right;

* TreeNode(int x) : val(x), left(NULL), right(NULL) {}

* };

class Solution {

public:

vector<int> result;

vector<int> inorderTraversal(TreeNode* root) {

dfs(root);

return result;

}

void dfs(TreeNode* root) {

if(root == NULL) return ;

dfs(root->left);

result.push_back(root->val);

dfs(root->right);

}

};

❤ 点击这里 -> 订阅《PAT | 蓝桥 | LeetCode学习路径 & 刷题经验》by 柳婼

❤ 点击这里 -> 订阅《从放弃C语言到使用C++刷算法的简明教程》by 柳婼

❤ 点击这里 -> 订阅PAT甲级乙级、蓝桥杯、GPLT天梯赛、LeetCode题解离线版