LeetCode –Binary Tree Preorder Traversal
题目: Given a binary tree, return the inorder traversal of its nodes’ values. Example: Input: 1 2 3 4 5 6 [1,null,2,3] 1 \ 2 / 3 Output: 1 [1,3,2] Follow up: Recursive solution is trivial, could you do it iteratively? 解题: 遍历顺序“根-左-右” 实现: 使用栈实现 从根节点开始,先将根节点压入栈,保存结果值,然后移动到左节点, 保证“根-左”顺序, 为空后出栈,取其右节点入栈中。这样就保证了访问顺序为“根-左-右”。 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 /** * 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> preorderTraversal(TreeNode* root) { vector<int> res; stack<TreeNode*> ss; TreeNode *p = root; while (!ss.empty()||p){ if (p!=NULL){ ss.push(p); res.push_back(p->val); p = p->left; } else{ p = ss.top(); ss.pop(); p=p->right; } } return res; } }; 实现: 递归实现 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> res; preorder(root, res); return res; } void preorder(TreeNode *root, vector<int> &res){ if (!root) return; res.push_back(root->val); preorder(root->left, res); preorder(root->right, res); } }; 参考: [LeetCode] 144. Binary Tree Preorder Traversal 二叉树的先序遍历 ...