题目描述
求给定的二叉树的后序遍历。
例如:
给定的二叉树为{1,#,2,3},
1↵ ↵ 2↵ /↵ 3↵
返回[3,2,1].
备注;用递归来解这道题太没有新意了,可以给出迭代的解法么?
Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree{1,#,2,3},
1↵ ↵ 2↵ /↵ 3↵
return[3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
示例1
输出
复制[3,2,1]
/**
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
class Solution {
public:
/**
*
* @param root TreeNode类
* @return int整型vector
*/
vector<int> postorderTraversal(TreeNode* root) {
// write code here
vector <int> res;
if (!root)
return res;
stack<TreeNode *> st;
st.push(root);
while (st.size())
{
TreeNode *temp=st.top();
st.pop();
res.push_back(temp->val);
if (temp->left)
st.push(temp->left);
if (temp->right)
st.push(temp->right);
}
reverse (res.begin(),res.end());
return res;
}
};