leetcode 513. Find Bottom Left Tree Value 最左边的值 + 一个简单的DFS深度优先遍历

Given a binary tree, find the leftmost value in the last row of the tree.

Example 1:

Input:

2
/ \
1   3

Output:

1

Example 2:

Input:

1
/ \
2   3
/   / \
4   5   6
/
7

Output:

7

Note: You may assume the tree (i.e., the given root node) is not NULL.

题意很简单，直接做DFS深度优先遍历即可，在遍历的时候作比较即可

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <cmath>

using namespace std;

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

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

int maxDepth = 0 , res = root->val;
getAll(root,maxDepth,0,res);
return res;
}

void getAll(TreeNode* root, int& maxDepth, int depth,int& res)
{
if (root == NULL)
return;
else
{
getAll(root->left,  maxDepth, depth + 1, res);
getAll(root->right, maxDepth, depth + 1, res);
if (depth > maxDepth)
{
maxDepth = depth;
res = root->val;
}
return;
}
}
};