LeetCode 104 Maximum Depth of Binary Tree DFS

Given the root of a binary tree, return its maximum depth.

A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Solution

很基础的 \(DFS\): \(\text{dfs}(root,depth)\)。当没有左右子节点时，直接返回；如果只有一个分支，那么在那个分支继续 \(\text{dfs}\)；如果有两个分支，则：

\[\max(\ \text{dfs}(root.left, depth+1),\text{dfs}(root.right,depth+1)\ ) \]

点击查看代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root) {
        if(!root)return 0;
        return dfs(root, 1);
    }
    int dfs(TreeNode* root, int depth){
        if(!root->left && !root->right)return depth;
        else if(root->left && !root->right)return dfs(root->left,depth+1);
        else if(root->right && !root->left)return dfs(root->right,depth+1);
        else return max(dfs(root->left,depth+1),dfs(root->right,depth+1));
    }
};