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617. 合并二叉树 #87

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Geekhyt opened this issue Sep 20, 2021 · 0 comments

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简单

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Geekhyt commented Sep 20, 2021

dfs 递归

在 root1 上直接修改，将两个树对应的节点相加后，赋值给 root1，然后递归执行两个左右子树。

const mergeTrees = function(root1, root2) {
    const preOrder = function(root1, root2) {
        if (!root1) return root2
        if (!root2) return root1
        root1.val += root2.val
        root1.left = preOrder(root1.left, root2.left)
        root1.right = preOrder(root1.right, root2.right)
        return root1
    }
    return preOrder(root1, root2)
}

时间复杂度: O(min(m,n))
空间复杂度: O(min(m,n))

Geekhyt added the 简单 label

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