leetcode [Binary Tree Level Order Traversal II]
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/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class Solution { public List<List<Integer>> levelOrderBottom(TreeNode root) { //采用BFS,一层一层的存起来,最后反序即可 List<List<Integer>> res = new ArrayList<>(); List<List<Integer>> resReal = new ArrayList<>(); List<Integer> tempRes; Queue<TreeNode> queue = new LinkedList<>(); TreeNode temp = null; if(root == null) return res; queue.add(root); while(!queue.isEmpty()){ tempRes = new ArrayList<>();//每一次都要申请一个新的空间 for(int i = 0, n = queue.size(); i < n; i++){ temp = queue.poll(); tempRes.add(temp.val);//存放每一层的tempRes if(temp.left != null) queue.add(temp.left); if(temp.right != null) queue.add(temp.right); } res.add(tempRes); //tempRes.clear();//放完后要清空,接着放下一层,这个不对,因为存入res的是tempRes这个堆空间,不是副本,若它变了res里的也会变 } for(int i = res.size() - 1; i >= 0; i--){ resReal.add(res.get(i)); } return resReal; }}
解法二(DFS):
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class Solution {void DFS(List<List<Integer>> res, TreeNode root, int level){//DFS里加一些其它的东西,用level记录深搜时结点的层数List<Integer> tempRes = new ArrayList<>();if(root == null) return;//递归出口//处理层数为level + 1的,因为List的下标是从0开始,所以根结点的level为0if(res.size() < level + 1){//第level+1层没有就建一个然后再添加该层的元素res.add(tempRes);res.get(level).add(root.val);}else{res.get(level).add(root.val);//第level+1层有就直接添加该层的元素}//递归调用,因为是从左到右,所以每一层添加也是按从左到右的顺序DFS(res, root.left, level + 1);DFS(res, root.right, level + 1);} public List<List<Integer>> levelOrderBottom(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); DFS(res, root, 0); List<List<Integer>> realRes = new ArrayList<>(); for(int i = res.size() - 1; i >=0; i--){ realRes.add(res.get(i)); } return realRes; }}
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