标签:
Given a binary tree, return the inorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3}
,
1 2 / 3
return [1,3,2]
.
Note: Recursive solution is trivial, could you do it iteratively?
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> ans; 14 if(!root) 15 return ans; 16 stack<TreeNode*> st; 17 TreeNode *rt=root;//现在要考虑的节点 18 while(rt||!st.empty()) 19 { 20 while(rt) 21 { 22 st.push(rt); 23 rt=rt->left; 24 } 25 TreeNode *temp=st.top(); 26 st.pop(); 27 ans.push_back(temp->val); 28 rt=temp->right; 29 } 30 return ans; 31 } 32 };
leetcode94 Binary Tree Inorder Traversal
标签:
原文地址:http://www.cnblogs.com/jsir2016bky/p/5106018.html