【leetcode刷题】T139-在每个树行中找最大值

示例：
输入: 
          1
         / \
        3   2
       / \   \  
      5   3   9 
输出: [1, 3, 9]

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def largestValues(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        # 层次遍历，找到最大值
        p = [root]
        q = []
        res = []
        val = root.val
        while p or q:
            if not p:
                p = copy.copy(q)
                q = []
                res.append(val)
                val = p[0].val
            cur = p.pop(0)
            val = max(val, cur.val)
            # 加入下一层节点
            if cur.left:
                q.append(cur.left)
            if cur.right:
                q.append(cur.right)
        # 最后一次循环，p、q皆为空
        res.append(val)
        return res

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> largestValues(TreeNode* root) {
        vector<int> res;
        if(!root)
            return res;
        queue<TreeNode*> p;
        queue<TreeNode*> q;
        p.push(root);
        int val = root->val;
        TreeNode* cur = NULL;
        while(!p.empty() || !q.empty()){
            if(p.empty()){
                while(!q.empty()){
                    p.push(q.front());
                    q.pop();
                }
                res.push_back(val);
                // 新的一层，改变初始值
                val = p.front()->val;
            }
            cur = p.front();
            p.pop();
            val = max(val, cur->val);
            if(cur->left)
                q.push(cur->left);
            if(cur->right)
                q.push(cur->right);
        }
        // 最后一次遍历，p、q都为空，未能添加最后一层的最大元素
        res.push_back(val);
        return res;
    }
};