144 Binary Tree Preorder traversal (binary trees pre-sequence traversal) + (binary tree, iteration)

translation

给定一个二叉树，返回其前序遍历的节点的值。例如：给定二叉树为 {1，#， 2， 3}   1         2    /   3返回 [123]备注：用递归是微不足道的，你可以用迭代来完成它吗？

Original

returntheofits nodes‘ values.For example:Given binary tree {1,#,2,3},   1         2    /   3return [1,2,3isit iteratively?

Analysis

The topic let me try to iterate, but the first to honestly give the recursion to write to say it ~

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

Next, let's write the iteration ...

/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode (int x): Val (x), left (null), right (null) {}*};*/classSolution { Public: vector<int>Preordertraversal (TreeNode *root) { vector<int>Result Stack<TreeNode*>Tempstack; while(!tempstack.empty () | | root! = NULL) {if(Root = NULL)                {Result.push_back (root->val);                Tempstack.push (root);            root = root->left; }Else{root = Tempstack.top ();                Tempstack.pop ();            root = root->right; }        }returnResult }};

144 Binary Tree Preorder traversal (binary trees pre-sequence traversal) + (binary tree, iteration)