这道题A的略烦
Binary Tree Level Order Traversal II
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] ]
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 int height; 12 TreeNode root; 13 14 public List<List<Integer>> levelOrderBottom(TreeNode root) { 15 this.height = getDepth(root);//获取树的高度 16 this.root = root; 17 18 List<List<Integer>> list = new ArrayList<List<Integer>>(); 19 for(int i = 0; i < height; i++){ 20 List<Integer> element = new ArrayList<Integer>(); 21 list.add(element); 22 }//this.height没有错 23 preOrder(root, list); 24 25 return list; 26 } 27 28 /** 29 * node在tree中高度 30 * @param root 31 * @param node 32 * @return 33 */ 34 public int getDepth(TreeNode root, TreeNode node){ 35 if(null == root) 36 return -Integer.MAX_VALUE; 37 if(node == root) 38 return 1; 39 else{ 40 int maxl = getDepth(root.left, node); 41 int maxr = getDepth(root.right, node); 42 if(maxl > maxr) 43 return maxl + 1; 44 else 45 return maxr + 1; 46 } 47 } 48 /** 49 * 获取树的高度 50 * @param root 51 * @return 52 */ 53 public int getDepth(TreeNode root){ 54 if(null == root){ 55 return 0; 56 }else{ 57 int maxl = getDepth(root.left); 58 int maxr = getDepth(root.right); 59 return maxl>maxr ? (maxl + 1):(maxr + 1); 60 } 61 } 62 /** 63 * 前序遍历树 64 * @param root 65 */ 66 public void preOrder(TreeNode root, List<List<Integer>> list){ 67 if(null != root){ 68 int height = getDepth(this.root, root); 69 List<Integer> element = list.get(this.height - height); 70 element.add(root.val); 71 preOrder(root.left, list); 72 preOrder(root.right, list); 73 } 74 } 75 }
ps:DFS实现
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 12 public List<List<Integer>> levelOrderBottom(TreeNode root) { 13 List<List<Integer>> container = new ArrayList<List<Integer>>(); 14 if(null == root){ 15 return container; 16 } 17 //List<TreeNode> theSameLevel = new ArrayList<TreeNode>();//存放同一层节点 18 Queue<TreeNode> theSameLevel = new LinkedList<TreeNode>(); 19 theSameLevel.add(root); 20 while(!theSameLevel.isEmpty()){//theSameLevel不为空 21 List<Integer> oneLevel = new ArrayList<Integer>(); 22 Queue<TreeNode> temp = new LinkedList<TreeNode>();//暂存同一层的结点 23 while(!theSameLevel.isEmpty()){ 24 TreeNode cur = theSameLevel.remove(); 25 oneLevel.add(cur.val); 26 if(null != cur.left) 27 temp.add(cur.left); 28 if(null != cur.right) 29 temp.add(cur.right); 30 } 31 theSameLevel = temp; 32 container.add(0, oneLevel); 33 } 34 35 return container; 36 37 } 38 39 40 }