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?
非递归实现树的前序遍历。前序遍历比较简单,注意入栈的时候应该按照右孩子节点在前的顺序,这样才能保证出栈顺序左先右后。
代码如下:
/** * 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) { vector<int> vi; if(root==NULL) return vi; stack<TreeNode*> st; st.push(root); while(!st.empty()) { root=st.top(); st.pop(); vi.push_back(root->val); if(root->right!=NULL) st.push(root->right); if(root->left!=NULL) st.push(root->left); } } };
leetcode 刷题之路 79 Binary Tree Preorder Traversal,布布扣,bubuko.com
leetcode 刷题之路 79 Binary Tree Preorder Traversal
原文:http://blog.csdn.net/u013140542/article/details/38515587