树的遍历

一、前序遍历
递归解法：

class Solution {public:    /**     * @param root: The root of binary tree.     * @return: Preorder in vector which contains node values.     */    vector preorderTraversal(TreeNode *root) {        vector val_vec;        traverse(root, val_vec);        return val_vec;    }private:    void traverse(TreeNode *root, vector &ret) {        if (root == NULL) {            return;        }        ret.push_back(root->val);        traverse(root->left, ret);        traverse(root->right, ret);    }};

非递归解法：

class Solution {public:    /**     * @param root: The root of binary tree.     * @return: Preorder in vector which contains node values.     */    vector preorderTraversal(TreeNode *root) {        vector val_vec;        stack s;        if (root == NULL) {            return val_vec;        }        s.push(root);        while (!s.empty()) {            const TreeNode *node = s.top();            s.pop();            val_vec.push_back(node->val);            if (node->right != NULL) {                s.push(node->right);            }            if (node->left != NULL) {                s.push(node->left);            }        }        return val_vec;    }};

二、中序遍历
递归解法：

class Solution {    /**     * @param root: The root of binary tree.     * @return: Inorder in vector which contains node values.     */public:    vector inorderTraversal(TreeNode *root) {        vector result;        traverse(root, result);        return result;    }private:    void traverse(TreeNode *root, vector &ret) {        if (root == NULL) {            return;        }        traverse(root->left, ret);        ret.push_back(root->val);        traverse(root->right, ret);    }};

非递归解法：

class Solution {    /**     * @param root: The root of binary tree.     * @return: Inorder in vector which contains node values.     */public:    vector inorderTraversal(TreeNode *root) {        vector result;        stack s;        while (!s.empty() || NULL != root) {            if (root) {                s.push(root);                root = root->left;            } else {                root = s.top();                s.pop();                result.push_back(root->val);                root = root->right;            }        }        return result;    }};

三、后序遍历
递归解法：

class Solution {    /**     * @param root: The root of binary tree.     * @return: Postorder in vector which contains node values.     */public:    vector postorderTraversal(TreeNode *root) {        vector result;        traverse(root, result);        return result;    }private:    void traverse(TreeNode *root, vector &ret) {        if (root == NULL) {            return;        }        traverse(root->left, ret);        traverse(root->right, ret);        ret.push_back(root->val);    }};

非递归解法：

class Solution {    /**     * @param root: The root of binary tree.     * @return: Postorder in vector which contains node values.     */public:    vector<int> postorderTraversal(TreeNode *root) {        // write your code here        stack<TreeNode*> s;        vector<int> res;        vector<bool> flag(1000);        int i = 0;        while (!s.empty() || root!=NULL) {            if (root) {                s.push(root);                flag[i++] = 0;                root = root->left;            } else if (flag[i-1]==0) {                root = s.top();                root = root->right;                flag[i-1] = 1;            } else {                TreeNode* tmp = s.top();                res.push_back(tmp->val);                s.pop();                i--;            }        }        return res;    }};