94. Binary Tree Inorder Traversal(Tree, stack)
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?
法I: recursion
/** * 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> inorderTraversal(TreeNode* root) { inorder(root); return ret; } void inorder(TreeNode* root){ if(root==NULL) return; inorder(root->left); ret.push_back(root->val); inorder(root->right); return; } private: vector<int> ret; };
法II:iteration
/** * 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> inorderTraversal(TreeNode* root) { if(root==NULL) return ret; TreeNode* current = root; stack<TreeNode*> s; s.push(current); while(1){ while(current->left){ s.push(current->left); //push left child current = current->left; flag.insert(current); } //After iterate left tree, visit root while(!s.empty() && s.top()->right == NULL){ current = s.top(); s.pop(); //pop root ret.push_back(current->val); //visit root } if(s.empty()) break; //terminate when stack is empty current = s.top(); s.pop(); //pop root ret.push_back(current->val); //visit root //go to right child s.push(current->right); //push right child current = current->right; } return ret; } private: vector<int> ret; };