java 解法, 执行用时: 0 ms, 内存消耗: 39.5 MB, 提交时间: 2023-09-06 09:39:14

/**
 * 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 TreeNode subtreeWithAllDeepest(TreeNode root) {
        return dfs(root).getValue();
    }

    private Pair<Integer, TreeNode> dfs(TreeNode node) {
        if (node == null)
            return new Pair<>(0, null);
        var left = dfs(node.left);
        var right = dfs(node.right);
        if (left.getKey() > right.getKey()) // 左子树更高
            return new Pair<>(left.getKey() + 1, left.getValue());
        if (left.getKey() < right.getKey()) // 右子树更高
            return new Pair<>(right.getKey() + 1, right.getValue());
        return new Pair<>(left.getKey() + 1, node); // 一样高
    }
}

javascript 解法, 执行用时: 68 ms, 内存消耗: 43.3 MB, 提交时间: 2023-09-06 09:38:45

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var subtreeWithAllDeepest = function(root) {
    return dfs(root)[1];
};

var dfs = function (node) {
    if (node === null)
        return [0, null];
    const [leftHeight, leftLca] = dfs(node.left);
    const [rightHeight, rightLca] = dfs(node.right);
    if (leftHeight > rightHeight) // 左子树更高
        return [leftHeight + 1, leftLca];
    if (leftHeight < rightHeight) // 右子树更高
        return [rightHeight + 1, rightLca];
    return [leftHeight + 1, node]; // 一样高
};

golang 解法, 执行用时: 0 ms, 内存消耗: 2.7 MB, 提交时间: 2023-09-06 09:37:50

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func subtreeWithAllDeepest(root *TreeNode) *TreeNode {
    _, lca := dfs(root)
    return lca
}

func dfs(node *TreeNode) (int, *TreeNode) {
    if node == nil {
        return 0, nil
    }
    leftHeight, leftLCA := dfs(node.Left)
    rightHeight, rightLCA := dfs(node.Right)
    if leftHeight > rightHeight { // 左子树更高
        return leftHeight + 1, leftLCA
    }
    if leftHeight < rightHeight { // 右子树更高
        return rightHeight + 1, rightLCA
    }
    return leftHeight + 1, node // 一样高
}

php 解法, 执行用时: 8 ms, 内存消耗: 19.2 MB, 提交时间: 2023-09-06 09:36:01

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     public $val = null;
 *     public $left = null;
 *     public $right = null;
 *     function __construct($val = 0, $left = null, $right = null) {
 *         $this->val = $val;
 *         $this->left = $left;
 *         $this->right = $right;
 *     }
 * }
 */
class Solution {

    /**
     * @param TreeNode $root
     * @return TreeNode
     */
    function subtreeWithAllDeepest($root) {
        return $this->dfs($root)[1];
    }
    
    function dfs($node) {
        if ( $node == null ) return [0, null];
        list($left_height, $left_lca) = $this->dfs($node->left);
        list($right_height, $right_lca) = $this->dfs($node->right);
        if ( $left_height > $right_height ) return [$left_height + 1, $left_lca];
        if ( $left_height < $right_height ) return [$right_height + 1, $right_lca];
        return [$left_height + 1, $node];
    }
}

python3 解法, 执行用时: 44 ms, 内存消耗: 16.4 MB, 提交时间: 2023-09-06 09:31:40

# 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 subtreeWithAllDeepest(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        def dfs(node: Optional[TreeNode]) -> (int, Optional[TreeNode]):
            if node is None:
                return 0, None
            left_height, left_lca = dfs(node.left)
            right_height, right_lca = dfs(node.right)
            if left_height > right_height:  # 左子树更高
                return left_height + 1, left_lca
            if left_height < right_height:  # 右子树更高
                return right_height + 1, right_lca
            return left_height + 1, node  # 一样高
        return dfs(root)[1]

python3 解法, 执行用时: 48 ms, 内存消耗: 15.2 MB, 提交时间: 2022-11-25 11:50:47

# 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(object):
    def subtreeWithAllDeepest(self, root):
        # The result of a subtree is:
        # Result.node: the largest depth node that is equal to or
        #              an ancestor of all the deepest nodes of this subtree.
        # Result.dist: the number of nodes in the path from the root
        #              of this subtree, to the deepest node in this subtree.
        Result = collections.namedtuple("Result", ("node", "dist"))
        def dfs(node):
            # Return the result of the subtree at this node.
            if not node: return Result(None, 0)
            L, R = dfs(node.left), dfs(node.right)
            if L.dist > R.dist: return Result(L.node, L.dist + 1)
            if L.dist < R.dist: return Result(R.node, R.dist + 1)
            return Result(node, L.dist + 1)

        return dfs(root).node

python3 解法, 执行用时: 44 ms, 内存消耗: 15.1 MB, 提交时间: 2022-11-25 11:50:20

# 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 subtreeWithAllDeepest(self, root: TreeNode) -> TreeNode:
        # Tag each node with it's depth.
        depth = {None: -1}
        def dfs(node, parent = None):
            if node:
                depth[node] = depth[parent] + 1
                dfs(node.left, node)
                dfs(node.right, node)
        dfs(root)

        max_depth = max(depth.values())

        def answer(node):
            # Return the answer for the subtree at node.
            if not node or depth.get(node, None) == max_depth:
                return node
            L, R = answer(node.left), answer(node.right)
            return node if L and R else L or R

        return answer(root)