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]
.
解法一:递归遍历
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> postorderTraversal(TreeNode* root) { 13 vector<int> result; 14 postorder(root,result); 15 return result; 16 17 18 } 19 20 void postorder(TreeNode *node,vector<int> &result) 21 { 22 if(node!=NULL) 23 { 24 postorder(node->left,result); 25 postorder(node->right,result); 26 result.push_back(node->val); 27 } 28 29 } 30 };