Same Tree

Same Tree

Structural recursion — both null OK; one null fails; equal values + recurse left and right.

Easy

Time O(n) Space O(n)

LeetCode

May 16, 2026 2 min read

Problem#

Return true iff two binary trees are structurally identical and have the same node values.

Examples#

[1,2,3], [1,2,3] → true.
[1,2], [1,null,2] → false; [1,2,1], [1,1,2] → false.

Constraints#

0 <= nodes <= 100.

Approach#

Recursion: both null → equal; one null → not equal; values differ → not equal; else recurse on children.

Solution#

struct TreeNode { int val; TreeNode* left; TreeNode* right; };

class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
        if (!p && !q) return true;
        if (!p || !q) return false;
        return p->val == q->val && isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
    }
};

func isSameTree(p *TreeNode, q *TreeNode) bool {
    if p == nil && q == nil {
        return true
    }
    if p == nil || q == nil {
        return false
    }
    return p.Val == q.Val && isSameTree(p.Left, q.Left) && isSameTree(p.Right, q.Right)
}

from typing import Optional

class Solution:
    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
        if not p and not q:
            return True
        if not p or not q:
            return False
        return p.val == q.val and self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)

function isSameTree(p, q) {
    if (!p && !q) return true;
    if (!p || !q) return false;
    return p.val === q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}

class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null) return true;
        if (p == null || q == null) return false;
        return p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    }
}

function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean {
    if (p === null && q === null) return true;
    if (p === null || q === null) return false;
    return p.val === q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}

Editorial#

Three cases at every recursive call cover all possibilities for the pair. The order of the children matters — (left, left) and (right, right), not crosswise.

Complexity#

Time: O(n).
Space: O(h) recursion (worst-case O(n)).

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