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

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.

The leetcode link

Pre-order DFS

The trick is using a list to track the following info,

List<Integer, Integer> -- <max_level_met, first_qualified_node_value>

Time complexity: O(n)
Runtime: 7ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int[] ret = new int[2];
    public int findBottomLeftValue(TreeNode root) {
        dfs(root, 1);
        return ret[1];
    }
    
    private void dfs(TreeNode root, int level){
        if (root == null) return;
        if (level > ret[0] && root.left == null && root.right == null) {
            ret[0] = level;
            ret[1] = root.val;
            return;
        }
        dfs(root.left, level+1);
        dfs(root.right, level+1);  
    }
}

imgRuntime: 7ms