513. 找树左下角的值

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 

解法

方法一：BFS

BFS 找最后一层第一个节点。

Python3JavaC++GoTypeScriptRust

# 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.
 * 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;
    }
}

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

/**
 * 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::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
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
    }
}

方法二：DFS

DFS 先序遍历，找深度最大的，且第一次被遍历到的节点。

Python3JavaC++GoTypeScriptRust

# 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:
        def dfs(root, curr):
            if root is None:
                return
            dfs(root.left, curr + 1)
            dfs(root.right, curr + 1)
            nonlocal ans, mx
            if mx < curr:
                mx = curr
                ans = root.val

        ans = mx = 0
        dfs(root, 1)
        return ans

/**
 * 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 ans = 0;
    private int mx = 0;

    public int findBottomLeftValue(TreeNode root) {
        dfs(root, 1);
        return ans;
    }

    private void dfs(TreeNode root, int curr) {
        if (root == null) {
            return;
        }
        dfs(root.left, curr + 1);
        dfs(root.right, curr + 1);
        if (mx < curr) {
            mx = curr;
            ans = root.val;
        }
    }
}

/**
 * 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 ans = 0;
    int mx = 0;
    int findBottomLeftValue(TreeNode* root) {
        dfs(root, 1);
        return ans;
    }

    void dfs(TreeNode* root, int curr) {
        if (!root) return;
        dfs(root->left, curr + 1);
        dfs(root->right, curr + 1);
        if (mx < curr) {
            mx = curr;
            ans = root->val;
        }
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findBottomLeftValue(root *TreeNode) int {
    ans, mx := 0, 0
    var dfs func(*TreeNode, int)
    dfs = func(root *TreeNode, curr int) {
        if root == nil {
            return
        }
        dfs(root.Left, curr+1)
        dfs(root.Right, curr+1)
        if mx < curr {
            mx = curr
            ans = root.Val
        }
    }
    dfs(root, 1)
    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 mx = 0;
    let ans = 0;

    function dfs(root, curr) {
        if (!root) {
            return;
        }
        dfs(root.left, curr + 1);
        dfs(root.right, curr + 1);
        if (mx < curr) {
            mx = curr;
            ans = root.val;
        }
    }
    dfs(root, 1);
    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::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, cur: i32, max: &mut i32, res: &mut i32) {
        if root.is_none() {
            return;
        }
        let root = root.as_ref().unwrap().borrow();
        Self::dfs(&root.left, cur + 1, max, res);
        Self::dfs(&root.right, cur + 1, max, res);
        if *max < cur {
            *max = cur;
            *res = root.val;
        }
    }

    pub fn find_bottom_left_value(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut max = 0;
        let mut res = 0;
        Self::dfs(&root, 1, &mut max, &mut res);
        res
    }
}

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