二叉树的最大深度和最小深度(编程实现)
1. 二叉树的最大深度:(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.
最大深度也是到叶子节点的长度,但是因为是求最大深度,单个孩子为空的非叶子节点不会干扰到结果,因此用最简洁的处理方式就可以搞定。
/** * 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; // 递归调用左子树 int rightDepth = maxDepth(root -> right) + 1; // 递归调用右子树 return max(leftDepth, rightDepth); // 返回左右子树中较大的一个, 注意最后不用加 1。 } };
2. 二叉树的最小深度(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.
因为深度是必须到叶子节点的距离,因此使用深度遍历时,不能单纯的比较左右子树的递归结果返回较小值,因为对于有单个孩子为空的节点,为空的孩子会返回0,但这个节点并非叶子节点,故返回的结果是错误的。因此,当发现当前处理的节点有单个孩子是空时,返回一个极大值INT_MAX,防止其干扰结果。
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)); } };