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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?
递归解决方案
class Solution { public: vector<int> res; void inorder(TreeNode* root){ if(root == NULL) return; inorder(root->left); res.push_back(root->val); inorder(root->right); } vector<int> inorderTraversal(TreeNode *root) { inorder(root); return res; } };
非递归解决方案
class Solution { public: vector<int> inorderTraversal(TreeNode *root) { vector<int> res; if(root == NULL) return res; stack<TreeNode *> nodeStack; TreeNode *current = root; while(!nodeStack.empty() || current ){ if(current != NULL){ nodeStack.push(current); current = current->left; }else{ current = nodeStack.top();nodeStack.pop(); res.push_back(current->val); current = current->right; } } return res; } };
Leetcode Binary Tree Inorder Traversal,布布扣,bubuko.com
Leetcode Binary Tree Inorder Traversal
标签：des style blog http color strong
原文地址：http://www.cnblogs.com/xiongqiangcs/p/3818925.html
