75 lines
1.7 KiB
C++
75 lines
1.7 KiB
C++
// Source : https://oj.leetcode.com/problems/binary-tree-inorder-traversal/
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// Author : Hao Chen
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// Date : 2014-06-27
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/**********************************************************************************
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*
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* Given a binary tree, return the inorder traversal of its nodes' values.
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*
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* For example:
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* Given binary tree {1,#,2,3},
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*
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* 1
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* \
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* 2
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* /
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* 3
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*
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* return [1,3,2].
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*
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* Note: Recursive solution is trivial, could you do it iteratively?
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*
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* confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
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*
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* OJ's Binary Tree Serialization:
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*
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* The serialization of a binary tree follows a level order traversal, where '#' signifies
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* a path terminator where no node exists below.
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*
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* Here's an example:
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*
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* 1
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* / \
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* 2 3
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* /
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* 4
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* \
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* 5
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*
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* The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
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*
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*
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**********************************************************************************/
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/**
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* Definition for binary tree
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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vector<int> inorderTraversal(TreeNode *root) {
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vector<TreeNode*> stack;
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vector<int> v;
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while(stack.size()>0 || root!=NULL){
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if (root!=NULL){
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stack.push_back(root);
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root = root->left;
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}else{
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if (stack.size()>0) {
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root = stack.back();
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stack.pop_back();
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v.push_back(root->val);
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root = root->right;
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}
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}
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}
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return v;
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}
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};
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