原题链接在这里：https://leetcode.com/problems/find-largest-value-in-each-tree-row/description/
题目：
You need to find the largest value in each row of a binary tree.
Example:
Input:
1
/
3 2
/
5 3 9
Output: [1, 3, 9]
题解：
可以采用BFS, 类似Binary Tree Level Order Traversal. 每层算最大值加入res中.
Time Complexity: O(n).
Space: O(n).
AC Java:
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 class Solution { 11 public List<Integer> largestValues(TreeNode root) { 12 List<Integer> res = new ArrayList<Integer>(); 13 if(root == null){ 14 return res; 15 } 16 17 LinkedList<TreeNode> que = new LinkedList<TreeNode>(); 18 que.add(root); 19 int curCount = 1; 20 int nextCount = 0; 21 int curLevelMax = Integer.MIN_VALUE; 22 while(!que.isEmpty()){ 23 TreeNode cur = que.poll(); 24 curCount--; 25 curLevelMax = Math.max(curLevelMax, cur.val); 26 27 if(cur.left != null){ 28 que.add(cur.left); 29 nextCount++; 30 } 31 32 if(cur.right != null){ 33 que.add(cur.right); 34 nextCount++; 35 } 36 37 if(curCount == 0){ 38 curCount = nextCount; 39 nextCount = 0; 40 res.add(curLevelMax); 41 curLevelMax = Integer.MIN_VALUE; 42 } 43 } 44 return res; 45 } 46 }
也可以DFS. 用depth来标记res中的index位置.
Time Complexity: O(n).
Space: O(logn). stack space.
AC Java:
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 class Solution { 11 public List<Integer> largestValues(TreeNode root) { 12 List<Integer> res = new ArrayList<Integer>(); 13 if(root == null){ 14 return res; 15 } 16 17 dfs(root, res, 0); 18 return res; 19 } 20 21 private void dfs(TreeNode root, List<Integer> res, int depth){ 22 if(root == null){ 23 return; 24 } 25 26 if(depth == res.size()){ 27 // 之前没有的碰到的深度. 28 res.add(root.val); 29 }else{ 30 // 之前有平级的深度. 31 res.set(depth, Math.max(res.get(depth), root.val)); 32 } 33 34 dfs(root.left, res, depth+1); 35 dfs(root.right, res, depth+1); 36 } 37 }