题目:
Given a binary tree, return the preorder traversal of its nodes‘ values.
Given binary tree {1,#,2,3},
1
2
/
3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
解答:
这题难在使用迭代而不是递归。思考一下,先序遍历过程中是不是处理根之后,根的左子树变成新的根,还有子树继续变成根……是不是很像一种数据结构:栈。
思路是这样:
/**
* 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> result;
result.clear();
if(root == NULL) return result;
stack<TreeNode*> right;
TreeNode* tmp = root;
bool tag = true;
while(tag || !right.empty())
{
tag = false;
while(tmp != NULL)
{
if(tmp->right != NULL)
{
right.push(tmp->right);
}
result.push_back(tmp->val);
tmp = tmp->left;
}
if(!right.empty())
{
tag = true;
tmp = right.top();
right.pop();
}
}
return result;
}
};代码有几个注意点:
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
vector<int> result;
stack<TreeNode*> s;
if (root) {
s.push(root);
}
while(!s.empty()) {
TreeNode* n = s.top();
s.pop();
while(n) {
if (n->right) {
s.push(n->right);
}
result.push_back(n->val);
n = n->left;
}
}
return result;
}
};【LeetCode从零单刷】Binary Tree Preorder Traversal
原文:http://blog.csdn.net/ironyoung/article/details/45015127