144. Binary Tree Preorder Traversal (二叉树前序遍历)

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree [1,null,2,3],

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

递归：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> res;
    void backtrack(TreeNode* root) {
        if(root==nullptr) {
            return;
        }
        res.push_back(root->val);
        backtrack(root->left);
        backtrack(root->right);
    }
    vector<int> preorderTraversal(TreeNode* root) {
       backtrack(root);
       return res;
    }
};

 1 class Solution {
 2     private List<Integer> res = new ArrayList<Integer>();
 3     public List<Integer> preorderTraversal(TreeNode root) {
 4         help(root);
 5         return res;
 6     }
 7     private void help(TreeNode root){
 8         if(root == null) return ;
 9         res.add(root.val);
10         help(root.left);
11         help(root.right);
12     }
13 }

非递归：

class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> stk;
        if(root != nullptr) stk.push(root);
        while(!stk.empty()) {
            TreeNode* currentnode = stk.top();
            stk.pop();
            res.push_back(currentnode->val);
            //至于为什么要右结点先入栈，是因为栈后进先出的特性。右结点先入栈，就会后输出右结点。
            if (currentnode->right !=nullptr) stk.push(currentnode->right);
            if (currentnode->left != nullptr) stk.push(currentnode->left);
        }
        return res;
    }
};

用stack 保存右节点跟左节点

 1 class Solution {
 2     public List<Integer> preorderTraversal(TreeNode root) {
 3         List<Integer> res = new ArrayList<Integer>();
 4         Stack<TreeNode> stack = new Stack();
 5         stack.push(root);
 6         while(!stack.isEmpty()){
 7             root = stack.pop();
 8             if (root!=null){
 9             res.add(root.val);
10             stack.push(root.right);
11             stack.push(root.left);
12             }
13         }
14         return res;
15     }
16 }