标签:递归后序遍历 postorder leetcode 非递归遍历
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?
方法一:递归遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> nums;
vector<int> postorderTraversal(TreeNode* root) {
postorder(root);
return nums;
}
void postorder(TreeNode * root){
if(root==NULL) return;
if(root->left) postorder(root->left);
if(root->right) postorder(root->right);
nums.push_back(root->val);
}
};
Leetcode[145]-Binary Tree Postorder Traversal
标签:递归后序遍历 postorder leetcode 非递归遍历
原文地址:http://blog.csdn.net/dream_angel_z/article/details/46480869