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Binary Tree Level Order Traversal II
题目描述:
Given a binary tree, return the bottom-up level order traversal of its nodes‘ values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree {3,9,20,#,#,15,7}
,
3 / 9 20 / 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
思路:先层序遍历二叉树,再将得到的结果翻转。
代码:
vector<vector<int> > levelOrderBottom(TreeNode *root) { vector<vector<int> > result; vector<int> sequence; TreeNode * flag = new TreeNode(0); queue<TreeNode *> treenode_queue; if(root == NULL) return result; treenode_queue.push(root); treenode_queue.push(flag); while(treenode_queue.size() >= 1) { TreeNode * node = treenode_queue.front(); treenode_queue.pop(); if(node == flag) { result.push_back(sequence); sequence.clear(); if(treenode_queue.empty()) break; treenode_queue.push(flag); } else { sequence.push_back(node->val); if(node->left != NULL) treenode_queue.push(node->left); if(node->right != NULL) treenode_queue.push(node->right); } } reverse(result.begin(),result.end()); return result; }
Binary Tree Level Order Traversal II
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