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@Geekhyt
Description
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))