Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a leaf value sequence.

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.

Example 1:

Input: root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8]
Output: true

Example 2:

Input: root1 = [1], root2 = [1]
Output: true

Example 3:

Input: root1 = [1], root2 = [2]
Output: false

Example 4:

Input: root1 = [1,2], root2 = [2,2]
Output: true

Example 5:

Input: root1 = [1,2,3], root2 = [1,3,2]
Output: false

Constraints:

  • The number of nodes in each tree will be in the range [1, 200].
  • Both of the given trees will have values in the range [0, 200].

Solution

Perform Depth-First Search on a tree and create an in-order list. Then, compare that list and the input list.

  • Use a stack to keep track of the DFS path
  • Check leaves 1 by 1, until the end or difference.

Time / space complexity: O(N)

bool leafSimilar(TreeNode* root1, TreeNode* root2) {
    stack<TreeNode*> s1 , s2;
    s1.push(root1); s2.push(root2);
    while (!s1.empty() && !s2.empty())
        if (dfs(s1) != dfs(s2)) return false;
    return s1.empty() && s2.empty();
}

int dfs(stack<TreeNode*>& s) {
    while (true) {
        TreeNode* node = s.top(); s.pop();
        if (node->right) s.push(node->right);
        if (node->left) s.push(node->left);
        if (!node->left && !node->right) return node->val;
    }
}
public boolean leafSimilar(TreeNode root1, TreeNode root2) {
    Stack<TreeNode> s1 = new Stack<>(), s2 = new Stack<>();
    s1.push(root1); s2.push(root2);
    while (!s1.empty() && !s2.empty())
        if (dfs(s1) != dfs(s2)) return false;
    return s1.empty() && s2.empty();
}

public int dfs(Stack<TreeNode> s) {
    while (true) {
        TreeNode node = s.pop();
        if (node.right != null) s.push(node.right);
        if (node.left != null) s.push(node.left);
        if (node.left == null && node.right == null) return node.val;
    }
}
def leafSimilar(self, root1, root2):
    def dfs(node):
        if not node: return
        if not node.left and not node.right: yield node.val
        for i in dfs(node.left): yield i
        for i in dfs(node.right): yield i
    return all(a == b for a, b in itertools.izip_longest(dfs(root1), dfs(root2)))

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