力扣——找数左下角的值

给定一个二叉树，在树的最后一行找到最左边的值。

示例 1:

输入:

    2
   / \
  1   3

输出:
1

示例 2:

输入:

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

输出:
7

注意: 您可以假设树（即给定的根节点）不为 NULL。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
     private int preLayer;
    private int value;

    public int findBottomLeftValue(TreeNode root) {
        preLayer = -1;
        traversal(root, 0);
        return value;
    }

    private void traversal(TreeNode node, int layer) {
        if (node == null) {
            return;
        }
        if (layer > preLayer) {
            preLayer = layer;
            value = node.val;
        }
        layer++;
        traversal(node.left, layer);
        traversal(node.right, layer);
    }
}

标签：node,layer,TreeNode,找数,preLayer,traversal,力扣,int,左下角
来源： https://www.cnblogs.com/JAYPARK/p/10526069.html