LeetCode: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.

算法1：dfs遞歸的求解

1 class Solution { 2 public: 3 int minDepth(TreeNode *root) { 4 // IMPORTANT: Please reset any member data you declared, as 5 // the same Solution instance will be reused for each test case. 6 if(root == NULL)return 0; 7 int res = INT_MAX; 8 dfs(root, 1, res); 9 return res; 10 } 11 void dfs(TreeNode *root, int depth, int &res) 12 { 13 if(root->left == NULL && root->right == NULL && res > depth) 14 {res = depth; return;} 15 if(root->left) 16 dfs(root->left, depth+1, res); 17 if(root->right) 18 dfs(root->right, depth+1, res); 19 } 20 }; View Code

還有一種更直觀的遞歸解法，分別求左右子樹的最小深度，然后返回左右子樹的最小深度中較小者+1

1 class Solution { 2 public: 3 int minDepth(TreeNode *root) { 4 if(root == NULL)return 0; 5 int minleft = minDepth(root->left); 6 int minright = minDepth(root->right); 7 if(minleft == 0) 8 return minright + 1; 9 else if(minright == 0) 10 return minleft + 1; 11 else return min(minleft, minright) + 1; 12 } 13 };

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算法2：層序遍歷二叉樹，找到最先遍歷到的葉子的層數就是樹的最小高度

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 */ 10 class Solution { 11 public: 12 int minDepth(TreeNode *root) { 13 // IMPORTANT: Please reset any member data you declared, as 14 // the same Solution instance will be reused for each test case. 15 //層序遍歷，碰到第一個葉子節點就停止,NULL作為每一層節點的分割標志 16 if(root == NULL)return 0; 17 int res = 0; 18 queue<TreeNode*> Q; 19 Q.push(root); 20 Q.push(NULL); 21 while(Q.empty() == false) 22 { 23 TreeNode *p = Q.front(); 24 Q.pop(); 25 if(p != NULL) 26 { 27 if(p->left)Q.push(p->left); 28 if(p->right)Q.push(p->right); 29 if(p->left == NULL && p->right == NULL) 30 { 31 res++; 32 break; 33 } 34 } 35 else 36 { 37 res++; 38 if(Q.empty() == false)Q.push(NULL); 39 } 40 } 41 return res; 42 } 43 };

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LeetCode: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.

算法1：dfs遞歸求解

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 */ 10 class Solution { 11 public: 12 int maxDepth(TreeNode *root) { 13 // IMPORTANT: Please reset any member data you declared, as 14 // the same Solution instance will be reused for each test case. 15 if(root == NULL)return 0; 16 int res = INT_MIN; 17 dfs(root, 1, res); 18 return res; 19 } 20 void dfs(TreeNode *root, int depth, int &res) 21 { 22 if(root->left == NULL && root->right == NULL && res < depth) 23 {res = depth; return;} 24 if(root->left) 25 dfs(root->left, depth+1, res); 26 if(root->right) 27 dfs(root->right, depth+1, res); 28 } 29 }; View Code

同上一題

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1 class Solution { 2 public: 3 int maxDepth(TreeNode *root) { 4 if(root == NULL)return 0; 5 int maxleft = maxDepth(root->left); 6 int maxright = maxDepth(root->right); 7 if(maxleft == 0) 8 return maxright + 1; 9 else if(maxright == 0) 10 return maxleft + 1; 11 else return max(maxleft, maxright) + 1; 12 } 13 };

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算法2：層序遍歷，樹的總層數就是樹的最大高度

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 */ 10 class Solution { 11 public: 12 int maxDepth(TreeNode *root) { 13 // IMPORTANT: Please reset any member data you declared, as 14 // the same Solution instance will be reused for each test case. 15 //層序遍歷計算樹的層數即可,NULL作為每一層節點的分割標志 16 if(root == NULL)return 0; 17 int res = 0; 18 queue<TreeNode*> Q; 19 Q.push(root); 20 Q.push(NULL); 21 while(Q.empty() == false) 22 { 23 TreeNode *p = Q.front(); 24 Q.pop(); 25 if(p != NULL) 26 { 27 if(p->left)Q.push(p->left); 28 if(p->right)Q.push(p->right); 29 } 30 else 31 { 32 res++; 33 if(Q.empty() == false)Q.push(NULL); 34 } 35 } 36 return res; 37 } 38 };

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