标签：des style blog io ar color sp for on
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?
分析：迭代版inorder traversal。对于任何一棵树，其最先被访问的是路径是最左侧的路径，在该最左侧路径上访问顺序呢是从叶子结点到根节点，很自然我们可以用一个stack保存这个路径。同时如果当前访问结点有右孩子，我们需把以右孩子为根的子树最左路径压入栈中。时间复杂度为O(n)，空间复杂度为O(h)。代码如下：
1 class Solution { 2 public: 3 vector<int> inorderTraversal(TreeNode *root) { 4 vector<int> result; 5 if(!root) return result; 6 7 stack<TreeNode*> S; 8 for(TreeNode *p = root; p; p = p->left) 9 S.push(p); 10 11 while(!S.empty()){ 12 TreeNode *tmp = S.top(); 13 S.pop(); 14 result.push_back(tmp->val); 15 for(TreeNode *p = tmp->right; p ; p = p->left) 16 S.push(p); 17 } 18 19 return result; 20 } 21 };
Leetcode: Binary Tree Inorder Traversal
标签：des style blog io ar color sp for on
原文地址：http://www.cnblogs.com/Kai-Xing/p/4143756.html
