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剑指 Offer II 045. 二叉树最底层最左边的值

题目描述

给定一个二叉树的 根节点 root，请找出该二叉树的 最底层 最左边 节点的值。

假设二叉树中至少有一个节点。

 

示例 1:

输入: root = [2,1,3]
输出: 1

示例 2: 

输入: [1,2,3,4,null,5,6,null,null,7]
输出: 7

 

提示:

  • 二叉树的节点个数的范围是 [1,104]
  • -231 <= Node.val <= 231 - 1 

 

注意：本题与主站 513 题相同： https://leetcode.cn/problems/find-bottom-left-tree-value/

解法

方法一

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# 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: TreeNode) -> int:
        q = deque([root])
        ans = -1
        while q:
            n = len(q)
            for i in range(n):
                node = q.popleft()
                if i == 0:
                    ans = node.val
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
        return ans

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/**
 * 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 = -1;
        while (!q.isEmpty()) {
            int n = q.size();
            for (int i = 0; i < n; i++) {
                TreeNode node = q.poll();
                if (i == 0) {
                    ans = node.val;
                }
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
        }
        return ans;
    }
}

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/**
 * 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;
        q.push(root);
        int ans = -1;
        while (!q.empty()) {
            for (int i = 0, n = q.size(); i < n; ++i) {
                TreeNode* node = q.front();
                if (i == 0) ans = node->val;
                q.pop();
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return ans;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findBottomLeftValue(root *TreeNode) int {
    q := []*TreeNode{root}
    ans := -1
    for n := len(q); n > 0; n = len(q) {
        for i := 0; i < n; i++ {
            node := q[0]
            q = q[1:]
            if i == 0 {
                ans = node.Val
            }
            if node.Left != nil {
                q = append(q, node.Left)
            }
            if node.Right != nil {
                q = append(q, node.Right)
            }
        }
    }
    return ans
}

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