标签:
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> inorderTraversal(TreeNode* root) { 13 vector<int> res; 14 if(root == NULL) 15 return res; 16 stack<TreeNode *> node; 17 node.push(root); 18 TreeNode *p,*cur; 19 if(root->left != NULL) 20 { 21 node.push(root->left); 22 p = root->left; 23 } 24 else 25 { 26 res.push_back(root->val); 27 node.pop(); 28 if(root->right != NULL) 29 { 30 node.push(root->right); 31 p = root->right; 32 } 33 else return res; 34 } 35 while(!node.empty()) 36 { 37 while(p->left != NULL) 38 { 39 p = p->left; 40 node.push(p); 41 } 42 cur = node.top(); 43 node.pop(); 44 res.push_back(cur->val); 45 if(cur->right != NULL) 46 { 47 p = cur->right; 48 node.push(p); 49 } 50 } 51 return res; 52 } 53 };
标签:
原文地址:http://www.cnblogs.com/daocaorenblog/p/5518888.html