Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Analyse: The same as Minimum Depth of Binary Tree.
1. Recursion
Runtime: 8ms.
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int maxDepth(TreeNode* root) { 13 if(!root) return 0; 14 15 int leftDepth = maxDepth(root->left); 16 int rightDepth = maxDepth(root->right); 17 18 if(leftDepth == 0 && rightDepth == 0) return 1; 19 if(leftDepth == 0) leftDepth = INT_MIN; 20 if(rightDepth == 0) rightDepth = INT_MIN; 21 22 return 1 + max(leftDepth, rightDepth); 23 } 24 };
2. Iteration: Compute the number of level. Initialize the level to be 0, after dealing with the current level, then level++.
Runtime: 8ms.
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int maxDepth(TreeNode* root) { 13 if(!root) return 0; 14 15 queue<TreeNode* > qu; 16 qu.push(root); 17 int level = 0; 18 while(!qu.empty()){ 19 int n = qu.size(); 20 while(n--){ 21 TreeNode* current = qu.front(); 22 qu.pop(); 23 24 if(current->left) qu.push(current->left); 25 if(current->right) qu.push(current->right); 26 } 27 level++; 28 } 29 return level; 30 } 31 };