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feat: add solutions to lc problem: No.0404 #2607

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Apr 17, 2024
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feat: add solutions to lc problem: No.0404 #2607

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258 changes: 211 additions & 47 deletions solution/0400-0499/0404.Sum of Left Leaves/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -42,29 +42,36 @@

## 解法

### 方法一
### 方法一：递归

我们首先判断 `root` 是否为空，如果为空则返回 $0$。

否则，我们递归调用 `sumOfLeftLeaves` 函数计算 `root` 的右子树中所有左叶子之和，并将结果赋给答案变量 $ans$。然后我们判断 `root` 的左子节点是否存在，如果存在，我们判断其是否为叶子节点，如果是叶子节点，则将其值加到答案变量 $ans$ 中，否则我们递归调用 `sumOfLeftLeaves` 函数计算 `root` 的左子树中所有左叶子之和，并将结果加到答案变量 $ans$ 中。

最后返回答案即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。

<!-- tabs:start -->

```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None


# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def sumOfLeftLeaves(self, root: TreeNode) -> int:
def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
if root is None:
return 0
res = 0
if root.left and root.left.left is None and root.left.right is None:
res += root.left.val
res += self.sumOfLeftLeaves(root.left)
res += self.sumOfLeftLeaves(root.right)
return res
ans = self.sumOfLeftLeaves(root.right)
if root.left:
if root.left.left == root.left.right:
ans += root.left.val
else:
ans += self.sumOfLeftLeaves(root.left)
return ans
```

```java
Expand All @@ -74,21 +81,29 @@ class Solution:
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int sumOfLeftLeaves(TreeNode root) {
if (root == null) {
return 0;
}
int res = 0;
if (root.left != null && root.left.left == null && root.left.right == null) {
res += root.left.val;
int ans = sumOfLeftLeaves(root.right);
if (root.left != null) {
if (root.left.left == root.left.right) {
ans += root.left.val;
} else {
ans += sumOfLeftLeaves(root.left);
}
}
res += sumOfLeftLeaves(root.left);
res += sumOfLeftLeaves(root.right);
return res;
return ans;
}
}
```
Expand Down Expand Up @@ -137,13 +152,15 @@ func sumOfLeftLeaves(root *TreeNode) int {
if root == nil {
return 0
}
res := 0
if root.Left != nil && root.Left.Left == nil && root.Left.Right == nil {
res += root.Left.Val
ans := sumOfLeftLeaves(root.Right)
if root.Left != nil {
if root.Left.Left == root.Left.Right {
ans += root.Left.Val
} else {
ans += sumOfLeftLeaves(root.Left)
}
}
res += sumOfLeftLeaves(root.Left)
res += sumOfLeftLeaves(root.Right)
return res
return ans
}
```

Expand All @@ -162,22 +179,19 @@ func sumOfLeftLeaves(root *TreeNode) int {
* }
*/

const dfs = (root: TreeNode | null, isLeft: boolean) => {
function sumOfLeftLeaves(root: TreeNode | null): number {
if (!root) {
return 0;
}
const { val, left, right } = root;
if (!left && !right) {
if (isLeft) {
return val;
let ans = sumOfLeftLeaves(root.right);
if (root.left) {
if (root.left.left === root.left.right) {
ans += root.left.val;
} else {
ans += sumOfLeftLeaves(root.left);
}
return 0;
}
return dfs(left, true) + dfs(right, false);
};

function sumOfLeftLeaves(root: TreeNode | null): number {
return dfs(root, false);
return ans;
}
```

Expand Down Expand Up @@ -235,27 +249,107 @@ impl Solution {
* };
*/

int dfs(struct TreeNode* root, int isLeft) {
int sumOfLeftLeaves(struct TreeNode* root) {
if (!root) {
return 0;
}
if (!root->left && !root->right) {
return isLeft ? root->val : 0;
int ans = sumOfLeftLeaves(root->right);
if (root->left) {
if (!root->left->left && !root->left->right) {
ans += root->left->val;
} else {
ans += sumOfLeftLeaves(root->left);
}
}
return dfs(root->left, 1) + dfs(root->right, 0);
}

int sumOfLeftLeaves(struct TreeNode* root) {
return dfs(root, 0);
return ans;
}
```

<!-- tabs:end -->

### 方法二
### 方法二：栈

我们也可以将方法一的递归改为迭代，使用栈来模拟递归的过程。

与方法一类似，我们首先判断 `root` 是否为空，如果为空则返回 $0$。

否则，我们初始化答案变量 $ans$ 为 $0$，然后初始化栈 $stk$，将 `root` 加入栈中。

当栈不为空时，我们弹出栈顶元素 `root`，如果 `root` 的左子节点存在，我们判断其是否为叶子节点，如果是叶子节点，则将其值加到答案变量 $ans$ 中，否则我们将其左子节点加入栈中。然后我们判断 `root` 的右子节点是否存在，如果存在，我们将其加入栈中。

最后返回答案即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。

<!-- tabs:start -->

```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
if root is None:
return 0
ans = 0
stk = [root]
while stk:
root = stk.pop()
if root.left:
if root.left.left == root.left.right:
ans += root.left.val
else:
stk.append(root.left)
if root.right:
stk.append(root.right)
return ans
```

```java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int sumOfLeftLeaves(TreeNode root) {
if (root == null) {
return 0;
}
Deque<TreeNode> stk = new ArrayDeque<>();
stk.push(root);
int ans = 0;
while (!stk.isEmpty()) {
root = stk.pop();
if (root.left != null) {
if (root.left.left == root.left.right) {
ans += root.left.val;
} else {
stk.push(root.left);
}
}
if (root.right != null) {
stk.push(root.right);
}
}
return ans;
}
}
```

```cpp
/**
* Definition for a binary tree node.
Expand Down Expand Up @@ -294,6 +388,76 @@ public:
};
```

```go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func sumOfLeftLeaves(root *TreeNode) (ans int) {
if root == nil {
return 0
}
stk := []*TreeNode{root}
for len(stk) > 0 {
root = stk[len(stk)-1]
stk = stk[:len(stk)-1]
if root.Left != nil {
if root.Left.Left == root.Left.Right {
ans += root.Left.Val
} else {
stk = append(stk, root.Left)
}
}
if root.Right != nil {
stk = append(stk, root.Right)
}
}
return
}
```

```ts
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/

function sumOfLeftLeaves(root: TreeNode | null): number {
if (!root) {
return 0;
}
let ans = 0;
const stk: TreeNode[] = [root];
while (stk.length) {
const { left, right } = stk.pop()!;
if (left) {
if (left.left === left.right) {
ans += left.val;
} else {
stk.push(left);
}
}
if (right) {
stk.push(right);
}
}
return ans;
}
```

<!-- tabs:end -->

<!-- end -->
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