1. 二叉樹的最大深度:(LeetCode104)
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.
最大深度也是到葉子節點的長度,但是因為是求最大深度,單個孩子為空的非葉子節點不會干擾到結果,因此用最簡潔的處理方式就可以搞定。
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int maxDepth(TreeNode *root) { if(!root) return 0; int leftDepth = maxDepth(root -> left) + 1; // 遞歸調用左子樹 int rightDepth = maxDepth(root -> right) + 1; // 遞歸調用右子樹 return max(leftDepth, rightDepth); // 返回左右子樹中較大的一個, 注意最后不用加 1。 } };
2. 二叉樹的最小深度(LeetCode111)
Given a binary tree, find its minimum depth.The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
因為深度是必須到葉子節點的距離,因此使用深度遍歷時,不能單純的比較左右子樹的遞歸結果返回較小值,因為對於有單個孩子為空的節點,為空的孩子會返回0,但這個節點並非葉子節點,故返回的結果是錯誤的。因此,當發現當前處理的節點有單個孩子是空時,返回一個極大值INT_MAX,防止其干擾結果。
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ class Solution { public: int minDepth(TreeNode *root) { if(!root) return 0; if(!root -> left && !root -> right) return 1; //Leaf means should return depth. int leftDepth = 1 + minDepth(root -> left); leftDepth = (leftDepth == 1 ? INT_MAX : leftDepth); int rightDepth = 1 + minDepth(root -> right); rightDepth = (rightDepth == 1 ? INT_MAX : rightDepth); //If only one child returns 1, means this is not leaf, it does not return depth. return min(leftDepth, rightDepth); } };
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int minDepth(TreeNode *root) { if (root == NULL) return 0; if (root->left == NULL && root->right == NULL) return 1; if (root->left == NULL) return minDepth(root->right) + 1; else if (root->right == NULL) return minDepth(root->left) + 1; else return 1 + min(minDepth(root->left), minDepth(root->right)); } };
