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513. Find Bottom Left Tree Value

Description

Given the root of a binary tree, return the leftmost value in the last row of the tree.

 

Example 1:

Input: root = [2,1,3]
Output: 1

Example 2:

Input: root = [1,2,3,4,null,5,6,null,null,7]
Output: 7

 

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • -231 <= Node.val <= 231 - 1

Solutions

  • /**
     * 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 findBottomLeftValue(TreeNode root) {
            Queue<TreeNode> q = new ArrayDeque<>();
            q.offer(root);
            int ans = 0;
            while (!q.isEmpty()) {
                ans = q.peek().val;
                for (int i = q.size(); i > 0; --i) {
                    TreeNode node = q.poll();
                    if (node.left != null) {
                        q.offer(node.left);
                    }
                    if (node.right != null) {
                        q.offer(node.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 findBottomLeftValue(TreeNode* root) {
            queue<TreeNode*> q{ {root} };
            int ans = 0;
            while (!q.empty()) {
                ans = q.front()->val;
                for (int i = q.size(); i; --i) {
                    TreeNode* node = q.front();
                    q.pop();
                    if (node->left) q.push(node->left);
                    if (node->right) q.push(node->right);
                }
            }
            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 findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
            q = deque([root])
            ans = 0
            while q:
                ans = q[0].val
                for _ in range(len(q)):
                    node = q.popleft()
                    if node.left:
                        q.append(node.left)
                    if node.right:
                        q.append(node.right)
            return ans
    
    
  • /**
     * Definition for a binary tree node.
     * type TreeNode struct {
     *     Val int
     *     Left *TreeNode
     *     Right *TreeNode
     * }
     */
    func findBottomLeftValue(root *TreeNode) int {
    	q := []*TreeNode{root}
    	ans := 0
    	for len(q) > 0 {
    		ans = q[0].Val
    		for i := len(q); i > 0; i-- {
    			node := q[0]
    			q = q[1:]
    			if node.Left != nil {
    				q = append(q, node.Left)
    			}
    			if node.Right != nil {
    				q = append(q, node.Right)
    			}
    		}
    	}
    	return ans
    }
    
  • /**
     * 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 findBottomLeftValue(root: TreeNode | null): number {
        let ans = 0;
        const q = [root];
        while (q.length) {
            ans = q[0].val;
            for (let i = q.length; i; --i) {
                const node = q.shift();
                if (node.left) {
                    q.push(node.left);
                }
                if (node.right) {
                    q.push(node.right);
                }
            }
        }
        return ans;
    }
    
    
  • // 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;
    use std::collections::VecDeque;
    impl Solution {
        pub fn find_bottom_left_value(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
            let mut queue = VecDeque::new();
            queue.push_back(root);
            let mut res = 0;
            while !queue.is_empty() {
                res = queue.front().unwrap().as_ref().unwrap().borrow_mut().val;
                for _ in 0..queue.len() {
                    let node = queue.pop_front().unwrap();
                    let mut node = node.as_ref().unwrap().borrow_mut();
                    if node.left.is_some() {
                        queue.push_back(node.left.take());
                    }
                    if node.right.is_some() {
                        queue.push_back(node.right.take());
                    }
                }
            }
            res
        }
    }
    
    

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