首页 > 代码库 > Leetcode dfs Binary Tree Postorder Traversal
Leetcode dfs Binary Tree Postorder Traversal
Binary Tree Postorder Traversal
Total Accepted: 28560 Total Submissions: 92333My SubmissionsGiven 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 iterativel
题意:后序遍历,不过给出了函数声明,限制了实现方式
思路:采用递归实现。因为函数声明是返回一个vector<int>,所以每个子树返回的是该子树的后序遍历的结果
按照 左、右、根的次序把根和左右子树的vector合并起来就可以了
/**
* 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> postorderTraversal(TreeNode *root) {
vector<int> post;
if(root == NULL) return post;
TreeNode *left = root->left;
TreeNode *right = root->right;
if(left) {
vector<int> left_vector = postorderTraversal(left);
post.insert(post.end(), left_vector.begin(), left_vector.end());
}
if(right){
vector<int> right_vector = postorderTraversal(right);
post.insert(post.end(), right_vector.begin(), right_vector.end());
}
post.push_back(root->val);
return post;
}
};
Leetcode dfs Binary Tree Postorder Traversal