//遞歸法
/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public int result;public int maxdepth=-1;//初始化為-1是因為當只有一個節點也就是只有根節點時，傳來的depth=0，0>-1，才會返回正確的左下角的值public void getresult(TreeNode root,int depth){if(root==null)return;if(root.left==null&&root.right==null){if(depth>maxdepth){maxdepth=depth;result=root.val;}}if(root.left!=null){depth++;getresult(root.left,depth);depth--;}if(root.right!=null){depth++;getresult(root.right,depth);depth--;}}public int findBottomLeftValue(TreeNode root) {getresult(root,0);return result;}
}

//迭代法
class Solution {public int findBottomLeftValue(TreeNode root) {Queue<TreeNode> que=new LinkedList<>();int res=0;que.offer(root);while(!que.isEmpty()){int size=que.size();for(int i=0;i<size;i++){TreeNode tmpnode=que.poll();if(i==0)res=tmpnode.val;//保存二叉樹最底邊第一個元素if(tmpnode.left!=null)que.offer(tmpnode.left);if(tmpnode.right!=null)que.offer(tmpnode.right);}}return res;}
}

513. 找樹左下角的值 - 力扣（LeetCode）513. 找樹左下角的值 - 給定一個二叉樹的 根節點 root，請找出該二叉樹的?最底層?最左邊?節點的值。假設二叉樹中至少有一個節點。?示例 1:[https://assets.leetcode.com/uploads/2020/12/14/tree1.jpg]輸入: root = [2,1,3]輸出: 1示例 2:[https://assets.leetcode.com/uploads/2020/12/14/tree2.jpg]輸入: [1,2,3,4,null,5,6,null,null,7]輸出: 7?提示: * 二叉樹的節點個數的范圍是 [1,104] * -231?<= Node.val <= 231?- 1?https://leetcode.cn/problems/find-bottom-left-tree-value