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]
.
题解:用了平凡的递归方法:
1 /** 2 * Definition for binary tree 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> answer; 14 post(root,answer); 15 return answer; 16 } 17 void post(TreeNode* root,vector<int>& nodes){ 18 if(root == NULL) 19 return; 20 post(root->left,nodes); 21 post(root->right,nodes); 22 nodes.push_back(root->val); 23 return; 24 } 25 };