LeetCode 515. Find Largest Value in Each Tree Row

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:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> largestValues(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null){
            return res;
        }
        
        LinkedList<TreeNode> que = new LinkedList<TreeNode>();
        que.add(root);
        int curCount = 1;
        int nextCount = 0;
        int curLevelMax = Integer.MIN_VALUE;
        while(!que.isEmpty()){
            TreeNode cur = que.poll();
            curCount--;
            curLevelMax = Math.max(curLevelMax, cur.val);
            
            if(cur.left != null){
                que.add(cur.left);
                nextCount++;
            }
            
            if(cur.right != null){
                que.add(cur.right);
                nextCount++;
            }
            
            if(curCount == 0){
                curCount = nextCount;
                nextCount = 0;
                res.add(curLevelMax);
                curLevelMax = Integer.MIN_VALUE;
            }
        }
        return res;
    }
}

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:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> largestValues(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null){
            return res;
        }
        
        dfs(root, res, 0);
        return res;
    }
    
    private void dfs(TreeNode root, List<Integer> res, int depth){
        if(root == null){
            return;
        }
        
        if(depth == res.size()){
            // 之前没有的碰到的深度.
            res.add(root.val);
        }else{
            // 之前有平级的深度.
            res.set(depth, Math.max(res.get(depth), root.val));
        }
        
        dfs(root.left, res, depth+1);
        dfs(root.right, res, depth+1);
    }
}