94、Binary Tree Inorder Traversal

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?

思路

中序遍历，先左，再根，后右，递归即可。

c++11

/**
 * 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> inorderTraversal(TreeNode* root) {
        vector<int> res;
        inorder(root, res);
        return res;
    }

    void inorder(TreeNode* root, vector<int> &res) {
        if (!root) return;
        if (root->left) inorder(root->left, res);
        res.emplace_back(root->val);
        if (root->right) inorder(root->right, res);
    }
};

ac结果：

Runtime: 8 ms, faster than 18.17% of C++ online submissions for Binary Tree Inorder Traversal.
Memory Usage: 9.3 MB, less than 50.80% of C++ online submissions for Binary Tree Inorder Traversal.