Minimum Depth of Binary Tree
OJ: https://oj.leetcode.com/problems/minimum-depth-of-binary-tree/
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.
思想:先序遍历。注意的是: 当只有一个孩子结点时,深度是此孩子结点深度加 1 .
/** * 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; int l = minDepth(root->left); int r = minDepth(root->right); return (l && r) ? min(l, r) + 1 : (l+r+1); } };
Balanced Binary Tree
OJ: https://oj.leetcode.com/problems/minimum-depth-of-binary-tree/
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
思想: 先序遍历。既要返回左右子树判断的结果,又要返回左右子树的深度进行再判断。
所以要么返回一个 pair<bool, int>, 要么函数参数增加一个引用来传递返回值。
方法1:返回一个 pair<bool, int>: (更简洁)
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ typedef pair<bool, int> Pair; Pair judge(TreeNode *root) { if(root == NULL) return Pair(true, 0); Pair L = judge(root->left); Pair R = judge(root->right); return Pair(L.first && R.first && abs(L.second-R.second) < 2, max(L.second, R.second)+1); } class Solution { public: bool isBalanced(TreeNode *root) { return judge(root).first; } };
方法二: 增加一个引用
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ bool judge(TreeNode *root, int& depth) { if(root == NULL) { depth = 0; return true; } int l, r; if(judge(root->left, l) && judge(root->right, r)) { depth = 1 + max(l, r); if(l-r <= 1 && l-r >= -1) return true; else return false; } } class Solution { public: bool isBalanced(TreeNode *root) { int depth; return judge(root, depth); } };
Maximum Depth of Binary Tree
OJ: https://oj.leetcode.com/problems/maximum-depth-of-binary-tree/
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.
注: 此题略水,于此带过。
class Solution { public: int maxDepth(TreeNode *root) { return root ? max(maxDepth(root->left), maxDepth(root->right))+1 : 0; } };