Binary Tree Level Order Traversal II leetcode java
2014-08-05 03:09
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题目: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
2 ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
3 if(root == null)
4 return res;
5
6 LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
7 queue.add(root);
8
9 int curLevCnt = 1;
int nextLevCnt = 0;ArrayList<Integer> levelres = new ArrayList<Integer>();while(!queue.isEmpty()){
TreeNode cur = queue.poll();
curLevCnt--;
levelres.add(cur.val);if(cur.left != null){
queue.add(cur.left);
nextLevCnt++;
}
if(cur.right != null){
queue.add(cur.right);
nextLevCnt++;
}if(curLevCnt == 0){
curLevCnt = nextLevCnt;
nextLevCnt = 0;
res.add(0,levelres); //insert one by one from the beginning
levelres = new ArrayList<Integer>();
}
}
return res;
}
{3,9,20,#,#,15,7},
3 / \ 9 20 / \ 15 7return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ] 题解: 这道题跟前面一道是一样的。。 只是存到res的结果顺序不一样罢了。 之前那个就是循序的存 这道题就是每得到一个行结果就存在res的0位置,这样自然就倒序了。 代码如下:1 public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
2 ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
3 if(root == null)
4 return res;
5
6 LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
7 queue.add(root);
8
9 int curLevCnt = 1;
int nextLevCnt = 0;ArrayList<Integer> levelres = new ArrayList<Integer>();while(!queue.isEmpty()){
TreeNode cur = queue.poll();
curLevCnt--;
levelres.add(cur.val);if(cur.left != null){
queue.add(cur.left);
nextLevCnt++;
}
if(cur.right != null){
queue.add(cur.right);
nextLevCnt++;
}if(curLevCnt == 0){
curLevCnt = nextLevCnt;
nextLevCnt = 0;
res.add(0,levelres); //insert one by one from the beginning
levelres = new ArrayList<Integer>();
}
}
return res;
}
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