Leetcode:maximum_depth_of_binary_tree题解

一、     题目

给定一个二叉树，求它的最大深度。最大深度是沿从根节点，到叶节点最长的路径。

二、     分析

(做到这里发现接连几道题都是用递归，可能就是因为自己时挑的简单的做的吧。)

找出最深路径，则每经过一个节点要加上1，当遇到空节点时，返回0。

在网上看了看有的做法除了麻烦点但是很不错的方法，比如使用栈和队列等

/**
 * 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==NULL)
        	return 0;
        return max(maxDepth(root->left),maxDepth(root->right))+1;
    }
};

队列

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
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;
    }

};</treenode*> 

栈

/**
 * 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 == 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;
    }
};
</treenode*> 

DFS
/**
 * 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 == NULL)return 0;
        int res = INT_MIN;
        dfs(root, 1, res);
        return res;
    }
    void dfs(TreeNode *root, int depth, int &res)
    {
        if(root->left == NULL && root->right == NULL && res < depth)
            {res = depth; return;}
        if(root->left)
            dfs(root->left, depth+1, res);
        if(root->right)
            dfs(root->right, depth+1, res);
    }
};

层次遍历

/**
 * 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) {
        //层次遍历树的层数,NULL为每一层节点的分割标志
        if(root == NULL)return 0;
        int res = 0;
        queue<TreeNode*> Q;
        Q.push(root);
        Q.push(NULL);
        while(Q.empty() == false)
        {
            TreeNode *p = Q.front();
            Q.pop();
            if(p != NULL)
            {
                if(p->left)Q.push(p->left);
                if(p->right)Q.push(p->right);
            }
            else
            {
                res++;
                if(Q.empty() == false)Q.push(NULL);
            }
        }
        return res;
    }
};