[LeetCode] Find Largest Value in Each Tree Row 找树每行最大的结点值

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]

这道题让我们找二叉树每行的最大的结点值，那么实际上最直接的方法就是用层序遍历，然后在每一层中找到最大值，加入结果res中即可，参见代码如下：

解法一：

class Solution {
public:vector<int> largestValues(TreeNode* root) {if (!root) return {};vector<int> res;queue<TreeNode*> q;q.push(root);while (!q.empty()) {int n = q.size(), mx = INT_MIN;for (int i = 0; i < n; ++i) {TreeNode *t = q.front(); q.pop();mx = max(mx, t->val);if (t->left) q.push(t->left);if (t->right) q.push(t->right);}res.push_back(mx);}return res;}
};

如果我们想用迭代的方法来解，可以用先序遍历，这样的话就需要维护一个深度变量depth，来记录当前结点的深度，如果当前深度大于结果res的长度，说明这个新一层，我们将当前结点值加入结果res中，如果不大于res的长度的话，我们用当前结点值和结果res中对应深度的那个结点值相比较，取较大值赋给结果res中的对应深度位置，参见代码如下：

解法二：

class Solution {
public:vector<int> largestValues(TreeNode* root) {if (!root) return {};vector<int> res;helper(root, 1, res);return res;}void helper(TreeNode* root, int depth, vector<int>& res) {if (depth > res.size()) res.push_back(root->val);else res[depth - 1] = max(res[depth - 1], root->val);if (root->left) helper(root->left, depth + 1, res);if (root->right) helper(root->right, depth + 1, res);}
};

参考资料：

https://discuss.leetcode.com/topic/79241/simple-and-easy-understand-c-dfs-solution

LeetCode All in One 题目讲解汇总(持续更新中...)

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