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Leetcode Find Bottom Left Tree Value problem solution

YASH PAL, 31 July 202422 January 2026

In this Leetcode Find Bottom Left Tree Value problem solution, we have given the root of a binary tree, return the leftmost value in the last row of the tree.

Leetcode Find Bottom Left Tree Value problem solution

Leetcode Find Bottom Left Tree Value problem solution in Python.

class Solution:
    def findBottomLeftValue(self, root: TreeNode) -> int:
        if not root:
            return 
        stack = [root]

        while stack:
            res = []
            temp = []
            for i in stack:
                if i:
                    if i.left:
                        temp.append(i.left)
                    if i.right:
                        temp.append(i.right)
                    res.append(i.val)
            stack = temp
                    
        return res[0]

Find Bottom Left Tree Value problem solution in Java.

class Solution {
    public int findBottomLeftValue(TreeNode root) {
        
        if(root == null) throw new IllegalArgumentException();
        
        TreeMap<Integer,Integer> depthsMap = new TreeMap();
        populateSiblings(root, depthsMap , 1);
        
        Integer last = depthsMap.lastKey();
        return depthsMap.get(last);
        
    }
    public void populateSiblings(TreeNode root, TreeMap<Integer,Integer> depthsMap , int depth )    
    {
        if (root == null) return;
        depthsMap.computeIfAbsent(depth , x -> root.val);
        populateSiblings(root.left, depthsMap , depth+1);
        populateSiblings(root.right, depthsMap , depth+1);
  
    }
}

Problem solution in C++.

class Solution {
public:
    int findBottomLeftValue(TreeNode* root) {
        queue<TreeNode*> q;
        q.push(root);
        
        int n = -1;
        while (q.size()) {
            n = q.front()->val;
            
            int s = q.size();
            for (int i = 0; i < s; ++i) {
                TreeNode* node = q.front();
                q.pop();
                if (node->left)  q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return n;
    }
};

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