LintCode (67) Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

Have you met this question in a real interview? 

Yes

Example

Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

Challenge

Can you do it without recursion?

真心败给这题了。。。真心只能靠背过了。。。。lol。。。再总结一下吧，希望下次碰到至少能知道。。。

模拟这个例子的思路的话，一路向左，一边向左一边push当前的位置， 1 -> stk. 

直到当前为Null, 1->null, 那么不能push到stk, 则回溯pop stack,打印当前结果， 去忘 stk.top()->right， 一路向左。。。

真正的inorder:

inorder(left)

cur

inorder(right)...

老感觉还是有一定理解上的距离的。。。。

不过之所以要先做这题是因为后面的iterator要用到它。。。

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/** * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; * } * } */ class Solution { /** * @param root: The root of binary tree. * @return: Inorder in vector which contains node values. */ public: vector<int> inorderTraversal(TreeNode *root) { // write your code here vector<int> result; stack<TreeNode*> stk; while(root || !stk.empty()){ if (root){ stk.push(root); root = root->left; } else{ root=stk.top(); stk.pop(); result.push_back(root->val); root=root->right; } } return result; } };