872 Leaf-Similar Trees 叶子相似的树
Description:
Consider all the leaves of a binary tree. From left to right order, the values of those leaves form a leaf value sequence.
binary tree
For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8).
Two binary trees are considered leaf-similar if their leaf value sequence is the same.
Return true if and only if the two given trees with head nodes root1 and root2 are leaf-similar.
Note:
Both of the given trees will have between 1 and 100 nodes.
题目描述:
请考虑一颗二叉树上所有的叶子，这些叶子的值按从左到右的顺序排列形成一个 叶值序列 。
二叉树
举个例子，如上图所示，给定一颗叶值序列为 (6, 7, 4, 9, 8) 的树。
如果有两颗二叉树的叶值序列是相同，那么我们就认为它们是 叶相似 的。
如果给定的两个头结点分别为 root1 和 root2 的树是叶相似的，则返回 true；否则返回 false 。
提示：
给定的两颗树可能会有 1 到 100 个结点。
思路:
- 迭代, 用栈记录
- 递归
找到所有叶子的值比较大小即可
时间复杂度O(n), 空间复杂度O(n), n是两棵树的最大结点数
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution
{
public:
bool leafSimilar(TreeNode* root1, TreeNode* root2)
{
return leaves(root1) == leaves(root2);
}
private:
vector<int> leaves(TreeNode* root)
{
stack<TreeNode*> s;
s.push(root);
vector<int> result;
while (s.size())
{
TreeNode* cur = s.top();
s.pop();
if (!cur -> left and !cur -> right) result.push_back(cur -> val);
if (cur -> left) s.push(cur -> left);
if (cur -> right) s.push(cur -> right);
}
return result;
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean leafSimilar(TreeNode root1, TreeNode root2) {
return leaves(root1, "").equals(leaves(root2, ""));
}
private String leaves(TreeNode root, String result) {
if (root == null) return result;
if (root.left == null && root.right == null) return result + ", " + root.val;
return leaves(root.left, result) + leaves(root.right, result);
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def leafSimilar(self, root1: TreeNode, root2: TreeNode) -> bool:
def dfs(root: TreeNode) -> List[int]:
s, result = [root], []
while s:
cur = s.pop()
if not cur.left and not cur.right:
result.append(cur.val)
if cur.left:
s.append(cur.left)
if cur.right:
s.append(cur.right)
return result
return dfs(root1) == dfs(root2)
# 官方题解
# def dfs(node):
# if node:
# if not node.left and not node.right:
# yield node.val
# yield from dfs(node.left)
# yield from dfs(node.right)
# return list(dfs(root1)) == list(dfs(root2))
