Commit 6990456

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add 515 solution for java

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	`1`	`+/**`
	`2`	`+ * Definition for a binary tree node.`
	`3`	`+ * public class TreeNode {`
	`4`	`+ * int val;`
	`5`	`+ * TreeNode left;`
	`6`	`+ * TreeNode right;`
	`7`	`+ * TreeNode(int x) { val = x; }`
	`8`	`+ * }`
	`9`	`+ */`
	`10`	`+public class Solution {`
	`11`	`+ // 深度遍历`
	`12`	`+ public List<Integer> largestValues(TreeNode root) {`
	`13`	`+ List<Integer> list = new ArrayList<>();`
	`14`	`+ dfs(list, root, 0);`
	`15`	`+ return list;`
	`16`	`+ }`
	`17`	`+`
	`18`	`+ private void dfs(List<Integer> list, TreeNode root, int level) {`
	`19`	`+ if (root == null) {`
	`20`	`+ return;`
	`21`	`+ }`
	`22`	`+ // 每深入一层,先把那一层的第一个节点加入返回 list中`
	`23`	`+ if (list.size() == level) {`
	`24`	`+ list.add(root.val);`
	`25`	`+ }`
	`26`	`+ // 此时 size > level ,那么就是开始遍历每一层的其他节点(不包括最左边的节点),`
	`27`	`+ // 直接比较list的对应下标(index)的值与当前值就好`
	`28`	`+ else {`
	`29`	`+ list.set(level, Math.max(list.get(level), root.val));`
	`30`	`+ }`
	`31`	`+ // 左右子树,深度要+1`
	`32`	`+ dfs(list, root.left, level + 1);`
	`33`	`+ dfs(list, root.right, level + 1);`
	`34`	`+ }`
	`35`	`+}`

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