[leetcode] 144. 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：

不管怎么样，recursive的方法最容易想到。

/** * 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;        if(!root) return res;        traveral(root,res);        return res;            }        void traveral(TreeNode* root, vector<int> &n){        n.push_back(root->val);        if(root->left) traveral(root->left,n);        if(root->right) traveral(root->right,n);    }};

解法二：

考虑是stack来辅助，储存root, right, left。如果left一直不为空，则一直先处理left。

/** * 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;        if(!root) return res;                stack<TreeNode*> q;        q.push(root);                while(!q.empty()){            TreeNode* node = q.top();            res.push_back(node->val);            q.pop();            if(node->right) q.push(node->right);            if(node->left) q.push(node->left);                    }        return res;    }};