[LeetCode] 100. Same Tree

Given the roots of two binary trees p and q, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.

Example 1:

Input: p = [1,2,3], q = [1,2,3]
Output: true

Example 2:

Input: p = [1,2], q = [1,null,2]
Output: false

Example 3:

Input: p = [1,2,1], q = [1,1,2]
Output: false

Constraints:
The number of nodes in both trees is in the range [0, 100].
-104 <= Node.val <= 104

相同的树。

给你两棵二叉树的根节点 p 和 q ，编写一个函数来检验这两棵树是否相同。 如果两个树在结构上相同，并且节点具有相同的值，则认为它们是相同的。

思路

题意是给两个树，判断他们是否相同。有两种思路，bfs和dfs。bfs是按层遍历tree，将每层的node塞进queue然后弹出比较，需要注意塞进queue的顺序。此处我只实现了dfs的做法。

复杂度

时间O(n)
空间O(n)

代码

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
	public boolean isSameTree(TreeNode p, TreeNode q) {
		if (p == null && q == null) {
			return true;
		}
		if (p == null || q == null) {
			return false;
		}
		if (p.val == q.val) {
			return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
		}
		return false;
	}
}

JavaScript实现

/**
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {boolean}
 */
const isSameTree = function(p, q) {
	// corner case
	if (p == null && q == null) {
		return true;
	}
	if (p == null || q == null) {
		return false;
	}

	// normal case
	if (p.val == q.val) {
		return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
	}
	return false;
};

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