标签：class efi lock solution des public node vector color
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].
递归
/** * 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> inorder; vector<int> inorderTraversal(TreeNode* root) { if (root == nullptr) return inorder; inorderTraversal(root->left); inorder.push_back(root->val); inorderTraversal(root->right); return inorder; } }; // 0 ms
迭代
/** * 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> inorder; vector<int> inorderTraversal(TreeNode* root) { if (root == nullptr) return inorder; stack<TreeNode*> stk; while (root != nullptr || !stk.empty()) { if (root != nullptr) { stk.push(root); root = root->left; } else { root = stk.top(); stk.pop(); inorder.push_back(root->val); root = root->right; } } return inorder; } }; // 3 ms
[LeetCode] Binary Tree Inorder Traversal
标签：class efi lock solution des public node vector color
原文地址：http://www.cnblogs.com/immjc/p/7500494.html
