105 lines
2.3 KiB
C++
105 lines
2.3 KiB
C++
// Source : https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/
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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 bottom-up level order traversal of its nodes' values.
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* (ie, from left to right, level by level from leaf to root).
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*
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* For example:
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* Given binary tree {3,9,20,#,#,15,7},
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*
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* 3
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* / \
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* 9 20
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* / \
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* 15 7
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*
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* return its bottom-up level order traversal as:
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*
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* [
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* [15,7],
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* [9,20],
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* [3]
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* ]
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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<vector<int> > levelOrderBottom(TreeNode *root) {
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queue<TreeNode*> q;
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vector< vector<int> > vv;
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vector<int> v;
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if (root){
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v.push_back(root->val);
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vv.push_back(v);
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}
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q.push(root);
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int i=0;
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vector<TreeNode*> vt;
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while(q.size()>0){
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TreeNode *p = q.front();
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q.pop();
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vt.push_back(p);
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if ( p==NULL ) {
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continue;
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}
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q.push(p->left);
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q.push(p->right);
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}
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int step = 2;
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int j;
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for (int i=1; i<vt.size(); i=j ){
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v.clear();
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int cnt=0;
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for (j=i; j<i+step && j<vt.size(); j++){
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if (vt[j]) {
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v.push_back(vt[j]->val);
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cnt += 2;
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}
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}
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step = cnt;
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if (v.size()>0) {
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vv.push_back(v);
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}
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}
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//reverse the order
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reverse(vv.begin(), vv.end());
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return vv;
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}
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};
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