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Binary Tree Preorder Traversal
Given a binary tree, return the preorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1
2
/
3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
解法一：递归
/** * Definition for binary tree * 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> ret; if(root == NULL) return ret; preOrder(root, ret); return ret; } void preOrder(TreeNode* root, vector<int> &ret) { if(root) { ret.push_back(root->val); preOrder(root->left, ret); preOrder(root->right, ret); } } };
解法二：借助栈的半层次遍历。
/** * Definition for binary tree * 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> ret; if(root == NULL) return ret; stack<TreeNode*> s; s.push(root); while(!s.empty()) { TreeNode* cur = s.top(); s.pop(); ret.push_back(cur->val); if(cur->right) s.push(cur->right); if(cur->left) s.push(cur->left); } return ret; } };
【LeetCode】Binary Tree Preorder Traversal (2 solutions)
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原文地址：http://www.cnblogs.com/ganganloveu/p/4097582.html
