Find Bottom Left Tree Value

• Initialize maxLevel = -1 and bottomLeftValue with a default value.
• Create a recursive helper function dfs(node, level).
• Start the traversal by calling dfs(root, 0).
• In dfs(node, level):
  • If node is null, return.
  • If the current level is greater than maxLevel, update maxLevel = level and bottomLeftValue = node.val.
  • Recursively call dfs(node.left, level + 1).
  • Recursively call dfs(node.right, level + 1).
• After the initial call completes, return bottomLeftValue.

The core idea is to traverse the tree while passing down the current level (or depth). We maintain two global or member variables: maxLevel to store the maximum depth reached, and bottomLeftValue to store the value of the leftmost node at that depth.\n\nWe define a helper function, dfs(node, level). The traversal order is crucial: we must visit the left subtree before the right one. This ensures that when we first encounter a new, deeper level, the node we are at is guaranteed to be the leftmost one for that level.\n\nInside the recursive function, we first check if the current level is greater than maxLevel. If it is, we've found a new deepest level. We update maxLevel to the current level and set bottomLeftValue to the current node's value. Then, we proceed with the recursive calls for the left and right children.\n\n```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 { private int maxLevel = -1; private int bottomLeftValue = 0;

  public int findBottomLeftValue(TreeNode root) { dfs(root, 0); return bottomLeftValue; }

  private void dfs(TreeNode node, int level) { if (node == null) { return; }

  1// If this is the first time we visit this level, it must be the leftmost node.2if (level > maxLevel) {3    maxLevel = level;4    bottomLeftValue = node.val;5}6 7// Traverse left first to ensure we see the leftmost node first at any level.8dfs(node.left, level + 1);9dfs(node.right, level + 1);

  } }