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?
分析:这道题与之前那道中序遍历是一样的题型,借助栈stack数据结构。先在stack中push进root,由于是前序遍历,故根结点的访问先于左结点和右结点,因此pop出一个结点,将它的值存入序列中,然后将它的右结点压入栈中,在压入它的左结点(这里为什么要先压入右结点呢?因为左结点要先于右结点访问,但是由于栈是后进先出,故要把左结点后于右结点压入栈中),直至栈为空位置。
/** * Definition for binary tree * 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) { stack<TreeNode*> StackTree; vector<int> data; if(root==NULL) return data; StackTree.push(root); while(!StackTree.empty()) { TreeNode *CurrentNode=StackTree.top(); StackTree.pop(); data.push_back(CurrentNode->val); if(CurrentNode->right!=NULL) StackTree.push(CurrentNode->right); if(CurrentNode->left!=NULL) StackTree.push(CurrentNode->left); } return data; } };
递归方法:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void Traversal(TreeNode *root,vector<int> &data) { if(root==NULL) return; data.push_back(root->val); Traversal(root->left,data); Traversal(root->right,data); } vector<int> preorderTraversal(TreeNode *root) { vector<int> data; Traversal(root,data); return data; } };