144. 二叉樹的前序遍歷(非遞迴)
阿新 • • 發佈:2021-06-28
144. 二叉樹的前序遍歷
難度簡單給你二叉樹的根節點root
,返回它節點值的前序遍歷。
示例 1:
輸入:root = [1,null,2,3] 輸出:[1,2,3]
示例 2:
輸入:root = [] 輸出:[]
示例 3:
輸入:root = [1] 輸出:[1]
示例 4:
輸入:root = [1,2] 輸出:[1,2]
示例 5:
輸入:root = [1,null,2] 輸出:[1,2]
提示:
- 樹中節點數目在範圍
[0, 100]
內 -100 <= Node.val <= 100
非遞迴方法:
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * };*/ class Solution { public: vector<int> preorderTraversal(TreeNode* root) { if(root == nullptr) return {}; stack<TreeNode* >s; vector<int> ans; while(root || !s.empty()){ while(root){ ans.push_back(root->val); s.push(root); root= root->left; } root = s.top(); s.pop(); root = root->right; } return ans; } };