[LeetCode]104.Maximum Depth of Binary Tree

【题目】

Maximum Depth of Binary Tree

 Total Accepted: 5260 Total
Submissions: 11532My Submissions

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.

【代码】

【递归】

时间复杂度为O(n) 空间复杂度为O（logn）

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

【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;
    }

};

【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) {
        // 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;
    }
};

【测试】

/*********************************
*   日期：2013-12-08
*   作者：SJF0115
*   题目: 104.Maximum Depth of Binary Tree
*   来源：http://oj.leetcode.com/problems/maximum-depth-of-binary-tree/
*   结果：AC
*   来源：LeetCode
*   总结：
**********************************/
#include <iostream>
#include <malloc.h>
#include <stdio.h>
using namespace std;

typedef struct TreeNode{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
}TreeNode,*BiTree;

//按先序序列创建二叉树
int CreateBiTree(BiTree &T){
	int data;
	//按先序次序输入二叉树中结点的值，‘-1’表示空树
	scanf("%d",&data);
	if(data == -1){
		T = NULL;
	}
	else{
		T = (BiTree)malloc(sizeof(TreeNode));
		//生成根结点
		T->val = data;
		//构造左子树
		CreateBiTree(T->left);
		//构造右子树
		CreateBiTree(T->right);
	}
	return 0;
}
//二叉树最大深度（递归）
int maxDepth(TreeNode *root) {
    if(root == NULL){
        return 0;
    }
    int left = maxDepth(root->left);
    int right = maxDepth(root->right);
    return 1 + max(left,right);
}

int main() {
    int i,n;
    BiTree T = NULL;
    CreateBiTree(T);
    printf("%d\n",maxDepth(T));
    return 0;
}