Solved by recursively traversing the given binary tree and maintaining “level” which will store the current node’s level in the tree.

  • 0
    class Solution {
        int findBottomLeftValue(TreeNode* root) 
              if (! root) return  0 ;
             int max_depth = 1 , res = root-> val;
            Helper (root, 1 , max_depth, res);
             return res;
         void Helper (TreeNode * node, int depth, int & max_depth, int & res) 
             if (! node) return ;
             if (depth> max_depth) 
                max_depth = depth;
                res = node-> val;
            Helper (node -> left, depth + 1 , max_depth, res);
            Helper (node -> right, depth + 1 , max_depth, res);

    If current node is left leaf, then check if its level is more than the level of deepest left leaf seen so far.
    -> If level is more then update the result.
    If current node is not leaf, then recursively find maximum depth in left and right subtrees, and return maximum of the two depths.

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