LeetCode | Maximum Depth of Binary Tree（二叉树的深度）

2023年11月12日 254次阅读

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.

题目解析：

求树的深度，很简单就能想到递归。但也有队列方法，和栈的方法，一一介绍。

方案一：

递归实现：

class Solution {
public:
    int maxDepth(TreeNode* root){
        if(root == NULL)
            return 0;

        int left = maxDepth(root->left);
        int right = maxDepth(root->right);
        int max = left > right? left : right;
        return max+1;
    }
};

方案二：

队列实现，跟前面两道题的意思一样。

class Solution {
public:
    //二叉树最大深度（层次遍历，遍历一层高度加1）
    int maxDepth(TreeNode *root) {
        int height = 0,rowCount = 1;
        if(root == NULL){
            return 0;
        }
        //创建队列
        queue<TreeNode *> queue;
        //添加根节点
        queue.push(root);
        //层次遍历
        while(!queue.empty()){
            //队列头元素
            TreeNode *node = queue.front();
            //出队列
            queue.pop();
            //一层的元素个数减1，一层遍历完高度加1
            rowCount --;
            if(node->left){
                queue.push(node->left);
            }
            if(node->right){
                queue.push(node->right);
            }
            //一层遍历完
            if(rowCount == 0){
                //高度加1
                height++;
                //下一层元素个数
                rowCount = queue.size();
            }
        }
        return height;
    }
 
};


方案三：

栈的实现，过程比较复杂，要仔细跟着流程走。

class Solution {
public:
    int maxDepth(TreeNode *root) {  
        // Start typing your C/C++ solution below  
        // DO NOT write int main() function  
        if(root == NULL) return 0;  
           
        stack<TreeNode*> S;  
           
        int maxDepth = 0;  
        TreeNode *prev = NULL;  
           
        S.push(root);  
        while (!S.empty()) {  
            TreeNode *curr = S.top();  
               
            if (prev == NULL || prev->left == curr || prev->right == curr) {  
                if (curr->left)  
                    S.push(curr->left);  
                else if (curr->right)  
                    S.push(curr->right);  
            } else if (curr->left == prev) {  
                if (curr->right)  
                    S.push(curr->right);  
            } else {  
                S.pop();  
            }  
            prev = curr;  
            if (S.size() > maxDepth)  
                maxDepth = S.size();  
        }  
        return maxDepth;  
    }  
};













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