leetcode 513. Find Bottom Left Tree Value 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;        }    }};