[LC] 513. Find Bottom Left Tree Value

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.

Solution 1:

DFS, preOrder

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int height = 0;
    int res = 0;
    public int findBottomLeftValue(TreeNode root) {
        helper(root, 1);
        return res;
    }
    
    private void helper(TreeNode root, int depth) {
        if (root == null) {
            return;
        }
        if (height < depth) {
            height = depth;
            res = root.val;
        }
        helper(root.left, depth + 1);
        helper(root.right, depth + 1);
    }
}

Solution 2:

BFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int findBottomLeftValue(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        int res = 0;
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            res = cur.val;
            if (cur.right != null) {
                queue.offer(cur.right);
            }
            if (cur.left != null) {
                queue.offer(cur.left);
            }
        }
        return res;
    }
}