Binary Tree Inorder Traversal--LeetCode

1.题目

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].
Note: Recursive solution is trivial, could you do it iteratively?

2.题意

二叉树中序遍历

3.分析

思路类似于Binary Tree Preorder Traversal
非递归解法借助栈，先将左孩子节点进栈，当左孩子节点为空，再将当前栈顶元素出栈，遍历其右子树

4.代码

1）递归版

class Solution {public:    vector<int> inorderTraversal(TreeNode* root) {        vector<int> result;        inorder(result, root);        return result;    }private:    void inorder(vector<int> &result, TreeNode *root) {        if(root == nullptr)            return;        inorder(result, root->left);        result.push_back(root->val);        inorder(result, root->right);    }};

2）迭代版

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