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?
一次通过~~ 耶
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<int> inorderTraversal(TreeNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function if(!root) return vector<int>(); vector<int> result; stack<pair<TreeNode*, bool>> s; s.push(pair<TreeNode*, bool>(root, false)); TreeNode *curNode; bool curFlag; while(!s.empty()){ curNode = s.top().first; curFlag = s.top().second; s.pop(); if(curFlag){ //second access result.push_back(curNode->val); }else{ //first access, push right, self, left if(curNode -> right) s.push(pair<TreeNode*, bool>(curNode->right, false)); s.push(pair<TreeNode*, bool>(curNode, true)); if(curNode -> left) s.push(pair<TreeNode*, bool>(curNode->left, false)); } } return result; } };