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.

class Solution {
    public int findBottomLeftValue(TreeNode root) {
        List<List<Integer>> help = levelOrder(root);
        return help.get(help.size() - 1).get(0);
    }
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        Helper(0, root, res);
        return res;
    }
    public void Helper(int height, TreeNode p, List<List<Integer>> res){
        if(p == null) return;
        if(height == res.size()){
            res.add(new ArrayList());
        }
        res.get(height).add(p.val);
        Helper(height + 1, p.left, res);
        Helper(height + 1, p.right, res);
    }
}

首先是我们的老朋友level order

public class Solution {
    int ans=0, h=0;
    public int findBottomLeftValue(TreeNode root) {
        findBottomLeftValue(root, 1);
        return ans;
    }
    public void findBottomLeftValue(TreeNode root, int depth) {
        if (h<depth) {ans=root.val;h=depth;}
        if (root.left!=null) findBottomLeftValue(root.left, depth+1);
        if (root.right!=null) findBottomLeftValue(root.right, depth+1);
    }
}
原文地址:https://www.cnblogs.com/wentiliangkaihua/p/12230207.html