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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].
inorder， 即按照“左根右”的顺序遍历，
1、当前访问节点不为空，则一直向左遍历，当前节点压栈
2、否则，取栈顶节点访问之，并遍历右节点
1 vector<int> inorderTraversal(TreeNode *root) 2 { 3 vector<int> tree_nodes; 4 stack<TreeNode*> stree; 5 TreeNode* visit = root; 6 7 while (!stree.empty() || (visit != NULL)) 8 { 9 if (visit != NULL) // if visit is not empty, push it into stack and visit left 10 { 11 stree.push(visit); 12 visit = visit->left; 13 } 14 else // visit is null, get top node from stack, visit it, and put visit to right 15 { 16 visit = stree.top(); 17 stree.pop(); 18 tree_nodes.push_back(visit->val); 19 visit = visit->right; 20 } 21 } 22 23 return tree_nodes; 24 }
leetcode 94. Binary Tree Inorder Traversal
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原文地址：http://www.cnblogs.com/ym65536/p/4231546.html
