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Well, this problem has the highest acceptance rate among all OJ problems. It has a very easy 1-line reursive solution. I am not sure whether this one can be called "DFS" so I only call it "recursive".
1 class Solution { 2 public: 3 int maxDepth(TreeNode* root) { 4 return root ? 1 + max(maxDepth(root -> left), maxDepth(root -> right)) : 0; 5 } 6 };
Well, you may also solve it using a level-order traversal (BFS) with a queue.
1 class Solution { 2 public: 3 int maxDepth(TreeNode* root) { 4 int depth = 0; 5 if (!root) return depth; 6 queue<TreeNode*> level; 7 level.push(root); 8 while (!level.empty()) { 9 depth++; 10 int n = level.size(); 11 for (int i = 0; i < n; i++) { 12 TreeNode* node = level.front(); 13 level.pop(); 14 if (node -> left) level.push(node -> left); 15 if (node -> right) level.push(node -> right); 16 } 17 } 18 return depth; 19 } 20 };
[LeetCode] Maximum Depth of Binary Tree
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4732902.html
