leetCode . Binary Tree Inorder/ Preorder/ Post Traversal iteratively

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问题

https://leetcode.com/problems/binary-tree-inorder-traversal/
https://leetcode.com/problems/binary-tree-postorder-traversal/
https://leetcode.com/problems/binary-tree-preorder-traversal/

解法

http://www.cnblogs.com/AnnieKim/archive/2013/06/15/morristraversal.html

Inorder 时间复杂度 3N 空间复杂度 O（1）

/** * 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> inorderTraversal(TreeNode* root) {        vector<int> ret;        TreeNode* cur = root;        while(cur != NULL)        {            if (cur->left == NULL)            {                ret.push_back(cur->val);                cur = cur->right;            }            else            {                TreeNode * pre = cur->left;                while(pre->right != NULL && pre->right != cur)                    pre = pre->right;                if (pre->right == NULL)                {                    pre->right = cur;                    cur = cur->left;                }                else                {                    pre->right = NULL;                    ret.push_back(cur->val);                    cur = cur->right;                }            }        }        return ret;    }};

Preorder 时间复杂度 3N 空间复杂度 O（1）

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

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