LintCode 二叉树的遍历 （非递归）

前序：

class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in vector which contains node values.
     */
    vector<int> preorderTraversal(TreeNode *root) {
        // write your code here
        stack<TreeNode*> s;
        vector<int> res;
        while (root!= nullptr || !s.empty()) {
            while (root != nullptr) {
                res.push_back(root->val);
                s.push(root);
                root = root->left;
            }
            if (!s.empty()) {
                root = s.top();
                s.pop();
                root = root->right;
            }
        }
        return res;
    }
};

中序：

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

兴许：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
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
        vector<int> res;
        stack<TreeNode*> s;
        TreeNode * cur;
        TreeNode *pre = nullptr;
        if (root == nullptr) {
            return res;
        }
        s.push(root);
        while (!s.empty()) {
            cur = s.top();
            if ((cur->left == nullptr && cur->right == nullptr) || (pre != nullptr && (pre==cur->left || pre == cur->right))) {
                res.push_back(cur->val);
                s.pop();
                pre = cur;
        }
        else {
            if (cur->right != nullptr) {
                s.push(cur->right);
            }
            if (cur->left != nullptr) {
                s.push(cur->left);
            }
            }
        }
        return res;
    }

};