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

Example 1:

Input:    2   / \  1   3Output:1 

Example 2: 

Input:        1       / \      2   3     /   / \    4   5   6       /      7Output:7

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

找出树左下角的值。

利用层次遍历和hashmap的key唯一性。

层次遍历二叉树，用hashmap来存储每一层的第一个节点，key为层数，value为节点值。

最后返回hashmap值为层数的那个value即可

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */static int layer = 0;class Solution {public:    int findBottomLeftValue(TreeNode* root) {        queue
     
       q;        q.push(root);        unordered_map
      
        m;        while (!q.empty()) {            int n = q.size();            layer++;            for (int i = 0; i < n; i++) {                TreeNode* curNode = q.front();                q.pop();                m.insert({layer, curNode});                if (curNode->left != nullptr)                    q.push(curNode->left);                if (curNode->right != nullptr)                    q.push(curNode->right);            }        }        return m[layer]->val;    }};// 13 ms
      
     

 Using dfs to solve this problem.

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int findBottomLeftValue(TreeNode* root) {        int val = 0;        int depth = 1;        int height = 0;        dfs(root, depth, height, val);        return val;    }        void dfs(TreeNode* node, int depth, int& height, int& val) {        if (node == nullptr)            return;        if (depth > height) {            val = node->val;            height = depth;        }        dfs(node->left, depth + 1, height, val);        dfs(node->right, depth + 1, height, val);    }};// 13 ms