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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?
中序遍历非递归版本
class Solution { private: vector<int> res; public: vector<int> inorderTraversal(TreeNode *root) { res.clear(); if(root==NULL) return res; stack<pair<TreeNode*,bool>> s; s.push(make_pair(root,false)); while (!s.empty()) { root=s.top().first; while (root!=NULL&&!s.top().second) { s.top().second=true; root=root->left; if(root!=NULL) s.push(make_pair(root,false)); } root=s.top().first->right; res.push_back(s.top().first->val); s.pop(); if(root!=NULL) s.push(make_pair(root,false)); } return res; } };
class Solution { private: vector<int> res; public: vector<int> inorderTraversal(TreeNode *root) { res.clear(); if(root==NULL) return res; stack<pair<TreeNode*,bool>> s; TreeNode* t; int used; s.push(make_pair(root,false)); while(!s.empty()) { t=s.top().first; used = s.top().second; s.pop(); if(!used) { if(t->right!=NULL) s.push( make_pair(t->right,false)); s.push(make_pair(t,true)); if(t->left!=NULL) s.push( make_pair(t->left,false)); } else res.push_back(t->val); } return res; } };
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原文地址：http://www.cnblogs.com/fightformylife/p/4244454.html
