标签：root otto nod 朋友 put most integer size inpu
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); } }
513. Find Bottom Left Tree Value
标签：root otto nod 朋友 put most integer size inpu
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/12230207.html
