100. Same Tree

Leetcode Tree Depth-first Search

Given two binary trees, 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:     1         1
          / \       / \
         2   3     2   3

        [1,2,3],   [1,2,3]

Output: true

Example 2:

Input:     1         1
          /           \
         2             2

        [1,2],     [1,null,2]

Output: false

Example 3:

Input:     1         1
          / \       / \
         2   1     1   2

        [1,2,1],   [1,1,2]

Output: false

分析

最初想到的就是一一比对二叉树的根节点、左节点和右节点，如果发现不想等的地方，则返回false；

public boolean isSameTree(TreeNode p, TreeNode q) {
    if (p == null || q == null) return p == q;
    Queue<TreeNode> queueP = new LinkedList<>(),
                    queueQ = new LinkedList<>();
    queueP.add(p); queueQ.add(q);

    while (!queueP.isEmpty() && !queueQ.isEmpty()) {
        if (queueP.peek().val != queueQ.peek().val)
            return false;

        TreeNode curP = queueP.poll(), curQ = queueQ.poll();
        if (curP.left == null) {
            if (curQ.left != null) return false;
        } else {
            queueP.add(curP.left);
            queueQ.add(curQ.left);
        }
        if (curP.right == null) {
            if (curQ.right != null) return false;
        }  else {
            queueP.add(curP.right);
            queueQ.add(curQ.right);
        }          
    }
    return queueP.isEmpty() && queueQ.isEmpty();
}

然后写了个递归版本：

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

递归真的是太简洁了！！！！而且运算时间还快！对于树的题目来说，还是写递归好。