[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?

递归    空间复杂度为O(n),时间复杂度为O(n)

/** * 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> res;    vector<int> preorderTraversal(TreeNode *root) {        if (!root) {            return res;        }        res.push_back(root->val);        preorderTraversal(root->left);        preorderTraversal(root->right);        return res;    }};

栈   空间复杂度为O(n),时间复杂度为O(n)

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

Morris二叉树遍历（参考http://www.it165.net/pro/html/201403/10857.html） 空间复杂度为O(1),时间复杂度为O(n)

/** * 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> res;        TreeNode *cur = root;        while (cur) {            if (cur->left == NULL) {                res.push_back(cur->val);                cur = cur->right;            } else {                TreeNode *node = cur->left;                while (node->right && node->right != cur) {                    node = node->right;                }                if (node->right == NULL) {                    res.push_back(cur->val);                    node->right = cur;                    cur = cur->left;                } else {                    node->right = NULL;                    cur = cur->right;                }            }        }        return res;    }};