Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1 2 / 3
return [1,3,2]
.
Note: Recursive solution is trivial, could you do it iteratively?
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
/** * Definition for binary tree * 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; if(root==NULL)return res; TreeNode *p=root; stack<TreeNode *> sta; while(p!=NULL||!sta.empty()) { while(p!=NULL) { sta.push(p); p=p->left; } if(!sta.empty()) { p=sta.top(); sta.pop(); res.push_back(p->val); p=p->right; } } return res; } /* void inorder(TreeNode *root, vector<int> &res) { if(root==NULL)return; inorder(root->left,res); res.push_back(root->val); inorder(root->right,res); return; } vector<int> inorderTraversal(TreeNode *root) { vector<int> res; inorder(root,res); return res; } */ };