代码题（6）— 二叉树的遍历

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

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

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

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

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

/**
 * 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> postorderTraversal(TreeNode* root) {
        vector<int> res;
        if(root == nullptr)
            return res;
        stack<TreeNode*> stNode;
        TreeNode* pre = nullptr;
        while(!stNode.empty() || root)
        {
            while(root)
            {
                stNode.push(root);
                root = root->left;
            }
            TreeNode* cur = stNode.top();
            if(cur->right == nullptr || pre == cur->right)  //如果没有右节点，或者右节点被访问过，则可以访问该节点
            {
                res.push_back(cur->val);
                stNode.pop();
                pre = cur;
            }
            else
                root = cur->right;
        }
        return res;
    }
};