leetcode 144. 二叉树的前序遍历(Binary Tree Preorder Traversal)

目录

• 题目描述：
• 示例:
• 解法：

题目描述：

给定一个二叉树，返回它的 前序 遍历。

示例:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [1,2,3]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

---

解法：

#define PR pair<TreeNode*, int> // node and its state

/**
 * 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:
    void inOrder(TreeNode* root, vector<int>& res){
        if(root != NULL){
            res.push_back(root->val);
            inOrder(root->left, res);
            inOrder(root->right, res);
        }
    }
    
    vector<int> inOrder(TreeNode* root){
        vector<int> res;
        if(root != NULL){
            stack<PR> stk;
            stk.push({root, 0});
            res.push_back(root->val);
            while(!stk.empty()){
                PR pr = stk.top();
                stk.pop();
                TreeNode* node = pr.first;
                int state = pr.second;
                if(state == 0){
                    // process left subtree
                    if(node->left == NULL){
                        stk.push({node, 1});
                    }else{
                        stk.push(pr);
                        res.push_back(node->left->val);
                        stk.push({node->left, 0});
                    }
                }else if(state == 1){
                    // process right subtree
                    if(node->right == NULL){
                        stk.push({node, 2});
                    }else{
                        stk.push(pr);
                        res.push_back(node->right->val);
                        stk.push({node->right, 0});
                    }
                }else{
                    if(!stk.empty()){
                        PR pr = stk.top();
                        stk.pop();
                        pr.second++;
                        stk.push(pr);
                    }
                }
            }
        }
        return res;
    }
    
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        
        // method 1:
        // inOrder(root, res);
        
        // method 2:
        res = inOrder(root);
        return res;
    }
};