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LeetCode 94:Binary Tree Inorder Traversal
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?
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//递归方法 class Solution { public: vector<int> res; vector<int> inorderTraversal(TreeNode* root) { tmpFunction(root); return res; } void tmpFunction(TreeNode* root){ if (root == NULL) return; tmpFunction(root->left); res.push_back(root->val); tmpFunction(root->right); } };
//非递归方法 class Solution { public: vector<int> inorderTraversal(TreeNode* root) { stack<TreeNode*> s; vector<int> res; if (root == NULL) return res; TreeNode* p=root; while (p!=NULL || !s.empty()) { while (p!=NULL) { s.push(p); p = p->left; //不断压入左孩子节点,直到节点为NULL } if (!s.empty()) { p = s.top(); s.pop(); res.push_back(p->val); p = p->right; } } return res; } };
LeetCode 94:Binary Tree Inorder Traversal
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