描述

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

Example 1:

Input: root = [1,null,2,3]
Output: [1,3,2]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> ans;
        inorder(root, ans);
        return ans;
    }
    void inorder(TreeNode* root, vector<int> &ans){
        if(root != nullptr){
            inorder(root->left, ans);
            ans.push_back(root->val);
            inorder(root->right, ans);
        }
    }
};

非递归

class Solution{
public:
	vector<int> inorderTraversal(TreeNode* root){
		vector<int> ans;
		stack<TreeNode *> stk;
		TreeNode* p = root;
		while(!stk.empty() || p!=nullptr){
			if(p != nullptr){
				stk.push(p);
				p = p->left;
			} else {
				p = stk.top();
				stk.pop();
				ans.push_back(p->val);
				p = p->right;
			}
		}
		return ans;
	}
}