import java.util.LinkedList;
import java.util.Queue;

/**

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

 Example 1:
 Input:

   2
  / \
 1   3

 Output:
 1

 Example 2:
 Input:

        1
       / \
     2   3
    /   / \
   4   5   6
  /
 7

 Output:
 7
 Note: You may assume the tree (i.e., the given root node) is not NULL.

 */
public class Find_Bottom_Left_Tree_Value {

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public int findBottomLeftValue(TreeNode root) {

            // bfs
            if (root == null) {
                return -1;
            }

            Queue<TreeNode> q = new LinkedList<>();
            q.offer(root);
            int currentLevelCount = 1;
            int nextLevelCount = 0;

            int result = root.val;

            boolean isLevelLeftMost = false;

            while (!q.isEmpty()) {
                TreeNode current = q.poll();
                currentLevelCount--;

                // check leftmost
                if (isLevelLeftMost) {
                    result = current.val;
                    isLevelLeftMost = false;
                }

                if (current.left != null) {
                    q.offer(current.left);
                    nextLevelCount++;
                }
                if (current.right != null) {
                    q.offer(current.right);
                    nextLevelCount++;
                }

                if (currentLevelCount == 0) {
                    currentLevelCount = nextLevelCount;
                    nextLevelCount = 0;

                    // mark next node of queue is the possible result
                    isLevelLeftMost = true;
                }
            }

            return result;
        }
    }
}

############

/**
 * 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;
    }
}

// OJ: https://leetcode.com/problems/find-bottom-left-tree-value/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int findBottomLeftValue(TreeNode* root) {
        int ans = 0;
        queue<TreeNode*> q;
        q.push(root);
        while (q.size()) {
            int cnt = q.size();
            ans = q.front()->val;
            while (cnt--) {
                root = q.front();
                q.pop();
                if (root->left) q.push(root->left);
                if (root->right) q.push(root->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.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def findBottomLeftValue(self, root):
    """
    :type root: TreeNode
    :rtype: int
    """

    def dfs(root, h, w):
      if not root:
        return (float("inf"), float("inf"), None)
      left = dfs(root.left, h - 1, w - 1)
      right = dfs(root.right, h - 1, w + 1)
      return min((h, w, root.val), left, right)

    return dfs(root, 0, 0)[2]

/**
 * 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
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int findBottomLeftValue(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        int bottomLeftValue = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            bottomLeftValue = queue.peek().val;
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                TreeNode left = node.left, right = node.right;
                if (left != null)
                    queue.offer(left);
                if (right != null)
                    queue.offer(right);
            }
        }
        return bottomLeftValue;
    }
}

############

/**
 * 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;
    }
}

// OJ: https://leetcode.com/problems/find-bottom-left-tree-value/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int findBottomLeftValue(TreeNode* root) {
        int ans = 0;
        queue<TreeNode*> q;
        q.push(root);
        while (q.size()) {
            int cnt = q.size();
            ans = q.front()->val;
            while (cnt--) {
                root = q.front();
                q.pop();
                if (root->left) q.push(root->left);
                if (root->right) q.push(root->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.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def findBottomLeftValue(self, root):
    """
    :type root: TreeNode
    :rtype: int
    """

    def dfs(root, h, w):
      if not root:
        return (float("inf"), float("inf"), None)
      left = dfs(root.left, h - 1, w - 1)
      right = dfs(root.right, h - 1, w + 1)
      return min((h, w, root.val), left, right)

    return dfs(root, 0, 0)[2]

/**
 * 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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