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404. Sum of Left Leaves

Description

Given the root of a binary tree, return the sum of all left leaves.

A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.

 

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 24
Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively.

Example 2:

Input: root = [1]
Output: 0

 

Constraints:

  • The number of nodes in the tree is in the range [1, 1000].
  • -1000 <= Node.val <= 1000

Solutions

  • /**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
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;
        }
        res += sumOfLeftLeaves(root.left);
        res += sumOfLeftLeaves(root.right);
        return res;
    }
}

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


class Solution:
    def sumOfLeftLeaves(self, root: 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


  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
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
	}
	res += sumOfLeftLeaves(root.Left)
	res += sumOfLeftLeaves(root.Right)
	return res
}

  • /**
 * 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)
 *     }
 * }
 */

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

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


  • // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, is_left: bool) -> i32 {
        if root.is_none() {
            return 0;
        }
        let node = root.as_ref().unwrap().borrow();
        let left = &node.left;
        let right = &node.right;
        if left.is_none() && right.is_none() {
            if is_left {
                return node.val;
            }
            return 0;
        }
        Self::dfs(left, true) + Self::dfs(right, false)
    }

    pub fn sum_of_left_leaves(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        Self::dfs(&root, false)
    }
}


  • /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (!root) {
            return 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 ans;
    }
};


  • /**
 * 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;
    }
}

  • /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (!root) {
            return 0;
        }
        int ans = 0;
        stack<TreeNode*> stk{{root}};
        while (!stk.empty()) {
            root = stk.top(), stk.pop();
            if (root->left) {
                if (!root->left->left && !root->left->right) {
                    ans += root->left->val;
                } else {
                    stk.push(root->left);
                }
            }
            if (root->right) {
                stk.push(root->right);
            }
        }
        return ans;
    }
};


  • /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

int sumOfLeftLeaves(struct TreeNode* root) {
    if (!root) {
        return 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 ans;
}

  • # 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


  • /**
 * 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
}

  • /**
 * 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;
}

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