算法：树相关算法题

算法：

1、判断一棵树是否是镜面树

思路：把树分为左右两棵树，要验证左右两棵树为镜面关系，那么左树的左子树，就和右树的右子树成镜面关系；左树的右子树和右树的左子树成镜面关系，递归判断到最后，如果全部都是镜面关系，那么该树就是镜面树，只要有两个节点不是镜面关系这棵树就不是镜面树。

// 测试链接：https://leetcode.com/problems/symmetric-tree
public class Code03_SymmetricTree {

	public static class TreeNode {
		public int val;
		public TreeNode left;
		public TreeNode right;
	}

	public static boolean isSymmetric(TreeNode root) {
		return isMirror(root, root);
	}

	public static boolean isMirror(TreeNode h1, TreeNode h2) {
		if (h1 == null ^ h2 == null) {
			return false;
		}
		if (h1 == null && h2 == null) {
			return true;
		}
		return h1.val == h2.val && isMirror(h1.left, h2.right) && isMirror(h1.right, h2.left);
	}

}

2、收集达标路径和

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

思路：
记录从根节点到当前节点的路径到path，路径和为preSum，遍历完把当前节点为根的子树后，把当前节点从路径中删除，每次遍历到叶子节点，都判断路径和是否和目标值相等，相等就存起来

/**
 * 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;
 *     }
 * }
 */
public class Solution{

	// 测试链接：https://leetcode.com/problems/path-sum-ii
	public class TreeNode {
		public int val;
		public TreeNode left;
		public TreeNode right;

		TreeNode(int val) {
			this.val = val;
		}
	}

	public List<List<Integer>> pathSum(TreeNode root, int sum) {
		List<List<Integer>> ans = new ArrayList<>();
		if (root == null) {
			return ans;
		}
		ArrayList<Integer> path = new ArrayList<>();
		process(root, path, 0, sum, ans);
		return ans;
	}

	public void process(TreeNode x, List<Integer> path, int preSum, int sum, List<List<Integer>> ans) {
		if (x.left == null && x.right == null) {
			if (preSum + x.val == sum) {
				path.add(x.val);
				ans.add(copy(path));
				path.remove(path.size() - 1);
			}
			return;
		}
		// x 非叶节点
		path.add(x.val);
		preSum += x.val;
		if (x.left != null) {
			process(x.left, path, preSum, sum, ans);
		}
		if (x.right != null) {
			process(x.right, path, preSum, sum, ans);
		}
		path.remove(path.size() - 1);
	}

	public List<Integer> copy(List<Integer> path) {
		List<Integer> ans = new ArrayList<>();
		for (Integer num : path) {
			ans.add(num);
		}
		return ans;
	}

}