标签：logs vector bsp rsa log via back ati code
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?
vector<int> inorderTraversal(TreeNode* root) { vector<int>result; stack<TreeNode*>st; while(root != NULL || !st.empty()) { if(root!= NULL) { st.push(root); root = root->left; } else { root = st.top(); result.push_back(root->val); st.pop(); root = root->right; } } return result; }
[leetcode-94-Binary Tree Inorder Traversal]
标签：logs vector bsp rsa log via back ati code
原文地址：http://www.cnblogs.com/hellowooorld/p/6665580.html
