二叉树深度高度节点数

• 104. Maximum Depth of Binary Tree
  • 思路
  • C++解法
  • Java解法
  • Python3解法
  • Go解法
• 111. Minimum Depth of Binary Tree
  • 思路
  • C++解法
  • Java解法
  • Python3解法
  • Go解法
• 222. Count Complete Tree Nodes
  • 思路
  • C++解法
  • Java解法
  • Python3解法
  • Go解法
• 110. Balanced Binary Tree
  • 思路
  • C++解法
  • Java解法
  • Python3解法
  • Go解法

104. Maximum Depth of Binary Tree

Given the root of a binary tree, return its maximum depth.

A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 3

Example 2:

Input: root = [1,null,2]
Output: 2

Constraints:

• The number of nodes in the tree is in the range [0, 10^4].
• -100 <= Node.val <= 100

思路

递归

C++解法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root) {
        if(root == NULL){
            return 0;
        }
        return max(maxDepth(root->left), maxDepth(root->right)) + 1;
    }
};

Java解法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int maxDepth(TreeNode root) {
        if(root == null) return 0;
        if(root.left == null && root.right == null) return 1;
        return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
    }
}

Python3解法

Go解法

111. 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.

Note: A leaf is a node with no children.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 2

Example 2:

Input: root = [2,null,3,null,4,null,5,null,6]
Output: 5

Constraints:

• The number of nodes in the tree is in the range [0, 10^5].
• -1000 <= Node.val <= 1000

思路

递归

C++解法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if(root == NULL){
            return 0;
        }
        if(root->left == NULL && root->right != NULL){
            return minDepth(root->right) + 1;
        }
        if(root->right == NULL && root->left != NULL){
            return minDepth(root->left) + 1;
        }
        return 1 + min(minDepth(root->left), minDepth(root->right));
    }
};

Java解法

Python3解法

Go解法

222. Count Complete Tree Nodes

Given the root of a complete binary tree, return the number of the nodes in the tree.

According to Wikipedia, every level, except possibly the last, is completely filled in a complete binary tree, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

Design an algorithm that runs in less than O(n) time complexity.

Example 1:

Input: root = [1,2,3,4,5,6]
Output: 6

Example 2:

Input: root = []
Output: 0

Example 3:

Input: root = [1]
Output: 1

Constraints:

• The number of nodes in the tree is in the range [0, 5 * 10^4].
• 0 <= Node.val <= 5 * 10^4
• The tree is guaranteed to be complete.

思路

递归，从根节点出发统计的节点数=从根节点左节点出发统计的节点数+从根节点右节点出发统计的节点数+1

C++解法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int countNodes(TreeNode* root) {
        if(root == NULL){
            return 0;
        }
        return countNodes(root->left) + countNodes(root->right) + 1;
    }
};

Java解法

Python3解法

Go解法

110. Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: true

Example 2:

Input: root = [1,2,2,3,3,null,null,4,4]
Output: false

Example 3:

Input: root = []
Output: true

Constraints:

• The number of nodes in the tree is in the range [0, 5000].
• -10^4 <= Node.val <= 10^4

思路

递归

C++解法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int getHeight(TreeNode* root){
        if(root == NULL){
            return 0;
        }
        int leftHeight = getHeight(root->left);
        if(leftHeight == -1){
            return -1;
        }
        int rightHeight = getHeight(root->right);
        if(rightHeight == -1){
            return -1;
        }
        if(abs(leftHeight - rightHeight) > 1){
            return -1;
        }else{
            return 1 + max(leftHeight, rightHeight);
        }
    }
    
    bool isBalanced(TreeNode* root) {
        return getHeight(root) == -1 ? false : true;
    }
};

Java解法

Python3解法

Go解法