标签：des style blog color os io ar strong for
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?
思路：1、非递归遍历 2、递归遍历
非递归：
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 11 struct node{ 12 TreeNode *root; 13 bool isFirst; 14 }; 15 16 class Solution { 17 public: 18 vector<int> result; 19 vector<int> postorderTraversal(TreeNode *root) { 20 vector<node> str; 21 if(root==NULL) return result; 22 TreeNode *p=root; 23 while(p!=NULL||str.size()!=0) 24 { 25 while(p!=NULL) 26 { 27 node n; 28 n.isFirst=true; 29 n.root=p; 30 str.push_back(n); 31 p=p->left; 32 } 33 if(str.size()!=0) 34 { 35 node temp=str.back(); 36 str.pop_back(); 37 38 if(temp.isFirst==true) 39 { 40 temp.isFirst=false; 41 str.push_back(temp); 42 p=temp.root->right; 43 } 44 else 45 { 46 result.push_back(temp.root->val); 47 p=NULL; 48 } 49 50 } 51 } 52 return result; 53 } 54 };
2：递归
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 11 class Solution { 12 public: 13 vector<int> result; 14 vector<int> postorderTraversal(TreeNode *root) { 15 digui(root); 16 return result; 17 } 18 19 void digui(TreeNode *root) 20 { 21 if(root!=NULL) 22 { 23 digui(root->left); 24 digui(root->right); 25 result.push_back(root->val); 26 } 27 } 28 };
Binary Tree Postorder Traversal <leetcode>
标签：des style blog color os io ar strong for
原文地址：http://www.cnblogs.com/sqxw/p/3960377.html
