【二叉树经典问题】 144. Binary Tree Preorder Traversal

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

For example:
Given binary tree {1,#,2,3},

   1    \     2    /   3

return [1,2,3].

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

递归非常简单：

/** * 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;        preorder(root,res);        return res;    }    void preorder(TreeNode* root, vector<int>& res){        if(!root) return;        res.push_back(root->val);        preorder(root->left,res);        preorder(root->right,res);    }};

非递归：

class Solution {public:    vector<int> preorderTraversal(TreeNode* root) {        vector<int> res;        stack<TreeNode*> s;        while(root||!s.empty()){            while(root){                res.push_back(root->val);                s.push(root);                root=root->left;            }            root=s.top();            s.pop();            root=root->right;        }        return res;    }};

还有一种非递归很直接：

class Solution {public:    vector<int> preorderTraversal(TreeNode *root) {        stack<TreeNode*> nodeStack;        vector<int> result;        //base case        if(root==NULL)        return result;        nodeStack.push(root);        while(!nodeStack.empty())        {            TreeNode* node= nodeStack.top();            result.push_back(node->val);            nodeStack.pop();            if(node->right)            nodeStack.push(node->right);            if(node->left)            nodeStack.push(node->left);        }        return result;            }};